Answer:
1). strong nuclear force 2). nuclear binding energy 3), mass defect
Explanation:
Right on Edge
1. Strong nuclear force the force that keeps the nucleons bound inside the nucleus of an atom.
2. Nuclear binding energy the amount of energy needed to split the nucleus into individual protons and neutrons.
3. Mass defect the difference between the mass of the nucleons and the mass of an Atom.
What is strong nuclear force ?The term strong nuclear force is defined as the force that binds protons and neutrons together. It also binds them all together in a nucleus and is responsible for the energy released in nuclear reactions.
The examples of strong nuclear force are the force that hold protons and neutrons in nuclei of atoms. The elements' greater than the hydrogen atom. The fusion of hydrogen into helium in the sun's core.
Thus, 1. option B, 2. option B and 3. option B is correct.
To learn more about the strong nuclear force, follow the link;
https://brainly.com/question/19271485
#SPJ2
According to the following reaction, what amount of Al2S3 remains when 20.00 g of Al2S3 and 2.00 g of H2O are reacted? A few of the molar masses are as follows: Al2S3 = 150.17 g/mol, H2O = 18.02 g/mol.
Answer:
Mass of Al2S3 remaining is 17.212 g
Explanation:
Equation of the reaction is given below:
Al2S3 + 6H2O -----> 2Al(OH)3 + 3H2S
From the balanced equation above
6 mole of H20 reacts with 1 mole of Al2S3
i.e. 6 * 18.02 g of H2O reacts with 1 * 150.71 g of Al2S3
= 108.12 g of H2O reacts with 150.71 g of Al2S3
Therefore 2.0 g of water will react with 2.0 * (150.71/108.12) g of Al2S3
= 2.788 g of Al2S3
Mass of Al2S3 remaining = 20.0 g - 2.788 g = 17.212 g
According to the properly balanced chemical equation, the amount (mass) of [tex]AL_2S_3[/tex] that remains after the chemical reaction is 17.22 grams.
Given the following data:
Mass of [tex]AL_2S_3[/tex] = 20.00 gramsMass of [tex]H_2O[/tex] = 2.00 gramsMolar mass of [tex]AL_2S_3[/tex] = 150.17 g/molMolar mass of [tex]H_2O[/tex] = 18.02 g/mol.To calculate the amount (mass) of [tex]AL_2S_3[/tex] that remains after the chemical reaction:
First of all, we would write a properly balanced chemical equation for this chemical reaction.
[tex]Al_2S_3 + 6H_2O ---> 2Al(OH)_3 + 3H_2S[/tex]
By stoichiometry:
1 mole of [tex]AL_2S_3[/tex] reacts with 6 moles of [tex]H_2O[/tex]
Next, we would calculate the mass of each compound.
For [tex]AL_2S_3[/tex]:
[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 1 \times 150.17[/tex]
Mass = 150.17 grams
For [tex]H_2O[/tex]:
[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 6 \times 18.02[/tex]
Mass = 108.12 grams
108.12 grams of [tex]H_2O[/tex] = 150.17 grams of [tex]AL_2S_3[/tex]
2.00 grams of [tex]H_2O[/tex] = X grams of [tex]AL_2S_3[/tex]
Cross-multiplying, we have:
[tex]108.12 \times X = 150.17 \times 2\\\\108.12X = 300.34\\\\X = \frac{300.34}{108.12}[/tex]
X = 2.78 grams of [tex]AL_2S_3[/tex]
Remaining mass = [tex]20.00 - 2.78[/tex]
Remaining mass = 17.22 grams of [tex]AL_2S_3[/tex]
Read more: https://brainly.com/question/13750908
Please help! (:
question above — how much money would you need to buy 7.0 lb of arugula? If 27lb of arugula cost $16
Answer:
$11.81
Explanation:
27 lb cost $16
27/16=$1.69 per pound
$1.69*7=$11.81 for 7 lbs
Could someone please help me with this chemistry question I will mark the correct answer as brainliest
Show work plzzz
Unknown Metal Bar #8
Mass of Unknown Metal bar 11.3g
Length of bar 13.90cm
Width of bar 2.9cm
Thickness of bar 0.081cm
1. Calculate the volume of the bar:
2. Calculate the (experimental) density of the bar:
3. Based on the provided list of (true) densities, what is the possible identity of the Unknown metal?
4. What is the percent difference between the true density of your metal and the calculated density?
= | − | ∗ 100%
Answer:
1= Volume
= Length x breath x height
= 13.90 x 2.9 x 0.081
=3.26511
2= Density = Mass ÷ volume
= 11.3 ÷ 3.26511
= 3.461 (3d.p)
idk the rest because you haven't shown a picture of the rest
Answer:
1. 3.3 cm³; 2. 3.5 g/cm³; 3. barium; 4. 4%
Explanation:
Experimental data:
Mass = 11.3 g
Length = 13.90 cm
Width = 2.9 cm
Thickness = 0.081 cm
Calculations:
1. Volume of bar
V = lwh = 13.90 cm × 2.9 cm × 0.081 cm = 3.3 cm³
2. Experimental density
[tex]\text{Density} = \dfrac{m}{V} = \dfrac{\text{11.3 g}}{\text{3.27 cm}^{3}} = \textbf{3.5 cm}^{\mathbf{3}}[/tex]
3. Identity of metal
The three most likely metals are scandium (3.00 g/cm³), barium (3.59 g/cm³), and yttrium (4.47 g/cm³)
The metal is probably barium.
4. Percent difference
[tex]\begin{array}{rcl}\text{Percent difference}&= &\dfrac{\lvert \text{ True - Calculated}\lvert}{ \text{True}} \times 100 \,\%\\\\& = & \dfrac{\lvert 3.59 - 3.5\lvert}{3.59} \times 100 \, \% \\\\& = & \dfrac{\lvert 0.1\lvert}{3.59} \times 100 \, \%\\ \\& = & 0.04 \times 100 \, \%\\& = & \mathbf{4 \, \%}\\\end{array}\\\text{The percent difference is $\large \boxed{\mathbf{4 \, \%} }$}[/tex]
What can be known about the salt sample that Gerry is looking at?
Answer:
That its small pointed. Pink(Himalayan salt)or white(normal salt)
Explanation:
Summa dees questions are so stupid, deys makin me salty.
A rule of thumb is that a reaction rate roughly doubles for every 10 °C increase in temperature. What is the activation energy of a reaction whose rate exactly doubles between 25.0 °C and 35.0 °C
Answer:
FOR EVERY 10 DEGREE CELSIUS INCREASE IN TEMPERATURE, THE ACTIVATION ENERGY THAT SHOWS THIS IS 52.4 KJ/MOL
Explanation:
From Arrhenius equation, the relationship between the rate constant and the temperature is as shown below:
k = Ae^ -Ea/RT
At initial temperature T1, the initial rate constant is (k1)
At final temperature T2, the final rate constant is k2
For the reaction rate to be doubled, we must double the rate constant which shows that the ratio of k2 / k1 must be equal to 2.
That is, k2 / k1 = 2 (rate is doubled)
Equating this into the Arrhenius equation, we have:
k2 / k1 = Ae^ (-Ea / R ) (1/ T2 - 1/T1)
2 = e^ (-Ea / R) (1 / T2 = 1 / T1)
Taking the natural logarithm of both sides:
ln 2 = - (Ea / R) (1 / T2 - 1 / T1)
Making Ea the subject of the formula, we obtain:
Ea = - (ln 2 R / (1 / T2- 1 / T1))
Let T1 = 25 C = 25 + 273 K = 298 K
T2 = 35 C = 35 + 273 K = 308 K
R = 8.314
So,
Ea = - (ln 2 * 8.314 / ( 1/308 - 1 / 298))
Ea = - (0.693 * 8.314 / 0.00324 - 0.00335)
Ea = - 5.7616 / -0.00011
Ea = 52 378,18 J / mol
So therefore, the activation energy Ea is 52.4 kJ/mol.
A 5.00-L tank contains helium gas at 1.50 atm. What is the pressure of the gas in mmHg
Answer:
1140 mmHg
Explanation:
1 atmosphere is 760 mmHg, so 1.5 atmospheres is ...
1.5×760 mmHg = 1140 mmHg
A gas has volume of 800.0mL at -23.0°c and 300.0torr. What would the volume of the gas be at 227.0°c and 600.0torr of pressure
Answer:
Explanation:
use gas law eqation
P1 * V1 / T1 = P2 * V2 /T2
600*V1/227 = 300*800/23
V1 = 300*800*227 / 23*600 = ............ can you solve this and get the answer?
Determine the limiting reactant in a mixture containing 95.7 g of B2O3, 75.7 g of C, and 369 g of Cl2. Calculate the maximum mass (in grams) of boron trichloride, BCl3, that can be produced in the reaction. The limiting reactant is:
Answer:
[tex]B_2O_3[/tex]
Explanation:
First, we have to find the reaction:
[tex]B_2O_3~+~C~+~Cl_2~->~BCl_3~+~CO[/tex]
The next step is to balance the reaction:
[tex]B_2O_3~+~3C~+~3Cl_2~->~2BCl_3~+~3CO[/tex]
Now, we have to calculate the molar mass for each compound, so:
[tex]B_2O_3=~69.62~g/mol[/tex]
[tex]C=~12~g/mol[/tex]
[tex]Cl_2=~70.96~g/mol[/tex]
With these values, we can calculate the moles of each compound:
[tex]95.7~g~B_2O_3\frac{1~mol~B_2O_3}{69.62~g~B_2O_3}=1.37~mol~B_2O_3[/tex]
[tex]75.7~g~C\frac{1~mol~C}{112~g~C}=6.30~mol~C[/tex]
[tex]369~g~Cl_2\frac{1~mol~Cl_2}{70.96~g~C}=5.20~mol~Cl_2[/tex]
Now we can divide by the coefficient of each compound in the balanced equation:
[tex]\frac{1.37~mol~B_2O_3}{1}=~1.37[/tex]
[tex]\frac{6.30~mol~C}{3}=~2.1[/tex]
[tex]\frac{5.20~mol~Cl_2}{3}=~1.73[/tex]
The smallest values are for [tex]B_2O_3[/tex], so this is our limiting reagent.
I hope it heps!
Give the IUPAC name for the following structure
Answer:
6-metyl-2-heptyne
Explanation:
C-C-C-C-C-C-C hept
2
C-C≡C-C-C-C-C 2-heptyne
C
| 6
C-C≡C-C-C-C-C
6-metyl-2-heptyne
The IUPAC name for the above structure is 6 methyl, hept-2-yne.
What is IUPAC?IUPAC stands for international Union of pure and applied chemistry. It is the body in charge of naming organic chemical compounds.
The naming is is based on a molecule's longest chain of carbons connected by single/double/triple bonds, whether in a continuous chain or in a ring etc.
According to this question, a structure is given. The following applies;
The compound has a triple bond located on the second carbon, hence, belongs to alkyne group. It has seven carbon atoms, hence, is heptyne. The methyl group is on the sixth carbon.Learn more about IUPAC at: https://brainly.com/question/33646537
#SPJ6
An equilibrium mixture of the three gases in a 1.00 L flask at 350 K contains 5.35×10-2 M CH2Cl2, 0.173 M CH4 and 0.173 M CCl4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.155 mol of CH4(g) is added to the flask?
Answer:
[CH₂Cl₂] = 7.07x10⁻² M
[CH₄] = 0.319 M
[CCl₄] = 0.164 M
Explanation:
The equilibrium reaction is the following:
2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)
The equilibrium constant of the above reaction is:
[tex] K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{0.173 M*0.173 M}{(5.35 \cdot 10^{-2} M)^{2}} = 10.5 [/tex]
When 0.155 mol of CH₄(g) is added to the flask we have the following concentration of CH₄:
[tex] C = \frac{\eta}{V} = \frac{0.155 mol}{1.00 L} = 0.155 M [/tex]
[tex]C_{CH_{4}} = 0.328 M[/tex]
Now, the concentrations at the equilibrium are:
2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)
5.35x10⁻² - 2x 0.328 + x 0.173 + x
[tex]K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{(0.328 + x)(0.173 + x)}{(5.35 \cdot 10^{-2} - 2x)^{2}}[/tex]
[tex]10.5*(5.35 \cdot 10^{-2} - 2x)^{2} - (0.328 + x)*(0.173 + x) = 0[/tex]
Solving the above equation for x:
x₁ = 0.076 and x₂ = -0.0086
Hence, the concentration of the three gases once equilibrium has been reestablished is:
[CH₂Cl₂] = 5.35x10⁻² - 2(-0.0086) = 7.07x10⁻² M
[CH₄] = 0.328 + (-0.0086) = 0.319 M
[CCl₄] = 0.173 + (-0.0086) = 0.164 M
We took x₂ value because the x₁ value gives a negative CH₂Cl₂ concentration.
I hope it helps you!
A certain substance X condenses at a temperature of 120.7 degree C. But if a 500, g sample of X is prepared with 55.4 g of urea (NH_2)_2 CO) dissolved in it, the sample is found to have a condensation point of 125.2 degree C instead. Calculate the molal boiling point elevation constant K_b of X. Round your answer to 2 significant digits.
Answer: The molal boiling point elevation constant [tex]k_b[/tex] of X is [tex]2.4^0C/m[/tex]
Explanation:
Formula used for Elevation in boiling point :
[tex]\Delta T_b=k_b\times m[/tex]
or,
[tex]T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]
where,
[tex]T_b-T^o_b =(125.2-120.7)^0C=4.5^0C[/tex]
[tex]k_b[/tex] = boiling point constant = ?
m = molality
[tex]w_2[/tex] = mass of solute (urea) = 55.4 g
[tex]w_1[/tex] = mass of solvent X = 500 g
[tex]M_2[/tex] = molar mass of solute (urea) = 60 g/mol
Now put all the given values in the above formula, we get:
[tex]4.5^oC=k_b\times \frac{55.4g\times 1000}{60\times 500g}[/tex]
[tex]k_b=2.4^0C/m[/tex]
Thus the molal boiling point elevation constant [tex]k_b[/tex] of X is [tex]2.4^0C/m[/tex]
Discuss any give ways by which
the falling moral standards of Ghanaian
youth can be minimised.
Answer:
The falling standards of Ghanaian youths can be minimized by proper upbringing of the children by their parents. The youths should be taught about what is wrong or right and there should be a corresponding reward for those who do good and exceptional in order to encourage others in towing that line and punishment should also be meted out to those who break the law. Mediocrity shouldn’t be celebrated and the elders should lead by example.
These will make the falling standards of Ghanaian youth get reduced.
A geochemist in the field takes a 46.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X. He notes the temperature of the pool, 21°C, and caps the sample carefully. Back in the lab, the geochemist filters the sample and then evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 0.87 g.
Required:
Using only the information above, can you calculate the solubility of X in water at 21°C? If yes, calculate it.
Answer: The solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.
Explanation:
The given data is as follows.
Volume of sample water = 46 ml
Temperature = [tex]21^oC[/tex]
After vaporization, washes and then drying the weight of mineral X = 0.87 g
This means that 46.0 ml of water contains 0.87 g of X. Therefore, grams present in 1 ml of water will be calculated as follows.
1 ml of water = [tex]\frac{0.87 g}{46.0 ml}[/tex]
= [tex]1.891 \times 10^{-2}[/tex] g/ml
Therefore, we can conclude that solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.
A compound D with the molecular formula C6H12 is optically inactive but can be resolved into enantiomers. On catalytic hydrogenation, D is converted to E (C6H14) and E is optically inactive. Propose structures for D and E. (Draw a three-dimensional formula for each using dashes and wedges around chiral centers.)
Answer:
D: CH2=CH-CH(CH3)-CH2-CH3 (R & S enantiomers)
E: CH3-CH2-(CH3)-CH2-CH3
(Please see the figures enclosed )
Explanation:
D is a racemic mixture (R & S) of 3-metyl-pent-1-ene, so it is optically inactive (although each of two enantiomers is optically active, the mixture is optically inactive. The reason is that two enantiomers are present in an equal amount).
E is optically inactive, so its structure has to be symmetric.
7.Which one of the following statements is not true?
1 point
O The molecules in a solid vibrate about a fixed position
O The molecules in a liquid are arranged in a regular pattern
The molecules in a gas exert negligibly small forces on each other, except during
collisions
The molecules of a gas occupy all the space available
Answer:
B. the molecules in liquid are loosely packed and scattered thus, they cannot be arranged
What would form a solution?
O A. Mixing two insoluble substances
O B. Mixing a solute and a solvent
O C. Mixing a solute and a precipitate
O D. Mixing two solutes together
Calculate the amount of ATP in kg that is turned over by a resting human every 24 hours. Assume that a typical human contains ~50g of ATP (Mr 505) and consumes ~8000 kJ of energy in food each day. The energy stored in the terminal anhydride bond of ATP under standard conditions is 30.6 kJmol-1. Assume also that the dietary energy is channeled through ATP with an energy transfer efficiency of ~50%.
Answer:
The correct answer is 66.35 kilograms.
Explanation:
Based on the data given in the question, the energy consumed by the body of a human being is 50%. Based on the given data, the energy consumed in a day is 8000 kJ, 50 percent is the energy transfer efficiency. Thus, the consumption of total energy is 4000 kJ, and for the transformation of ADP to ATP, the energy involved is 30.6 kJ per mole.
Hence, the total ATP produced in the process is,
ATP = 4000 kJ / 30.6 kJ/mol
= 130.7189 mol.
Thus, with the energy transfer efficiency of 50 percent, the total moles of ATP produced is 130.7 mol.
The mass of ATP can be calculated by using the formula,
moles = mass/molecular mass
The molecular mass of ATP is 507.18 g per mol
Now by putting the values we get,
mass of ATP = 130.7189 mol * 507.18 g/mol
= 66298.011 g or 66.298 kg
It is mentioned that human comprise 50 g of ATP or 0.05 kg of ATP. Therefore, the sum of the available ATP will be.
= Total production of ATP + Total ATP available
= 66.298 kg + 0.05 kg
= 66.348 kg
Hence, the sum of the ATP that is turned over by a resting human in a day is 66.35 kg.
Use the formation reactions below such that when added together, they match the balanced equation for the combustion of methane.
Cgraphite(s)+ 2H2(g) → CH4(g) ΔH 1=−74.80kJ
Cgraphite(s)+ O2(g) → CO2(g) ΔH2=−393.5k
H2(g)+ 1/2O2(g) → H2O(g) ΔH3=−241.80kJ
Calculate ΔHrxn for the combustion of methane, CH4(g).
CH4(g)+ 2O2(g) → CO2(g)+ 2H2O(g) ΔHrxn =--------------kJ
Answer:
ΔH of the reaction is -802.3kJ.
Explanation:
Using Hess's law, you can know ΔH of reaction by the sum of ΔH's of half-reactions.
Using the reactions:
(1) Cgraphite(s)+ 2H₂(g) → CH₄(g) ΔH₁ = −74.80kJ
(2) Cgraphite(s)+ O₂(g) → CO₂(g) ΔH₂ = −393.5k J
(3) H₂(g) + 1/2 O₂(g) → H₂O(g) ΔH₃ = −241.80kJ
The sum of (2) - (1) produce:
CH₄(g) + O₂(g) → CO₂(g) + 2H₂(g) ΔH' = -393.5kJ - (-74.80kJ) = -318.7kJ
And the sum of this reaction with 2×(3) produce:
CH₄(g) + 2 O₂(g) → CO₂(g) + 2H₂O(g) And ΔH = -318.7kJ + 2×(-241.80kJ) =
-802.3kJ
What are the relations between Electrochemistry and Cancer?
Answer: if im not wrong the relations are that the electrochemistry can detect the cancer and any other sickness
just like it does with chemical phenomena
=)
In E. coli, the enzyme hexokinase catalyzes the reaction: Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102. In the living E. coli cells, [ATP] = 7.9 mM; [ADP] = 1.04 mM, [glucose] = 2 mM, [glucose 6-phosphate] = 1 mM. Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.
Answer:
Explanation:
Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.
In the living E. coli cells,
[ATP] = 7.9 mM;
[ADP] = 1.04 mM,
[glucose] = 2 mM,
[glucose 6-phosphate] = 1 mM.
Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.
The reaction is given as
Glucose + ATP → glucose 6-phosphate + ADP
Now reaction quotient for given equation above is
[tex]q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}[/tex]
[tex]q=\frac{(1mm)\times (1.04 mm)}{(7.9mm)\times (2mm)} \\\\=6.582\times 10^{-2}[/tex]
so,
[tex]q<<K_e_q[/tex] ⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable until q = Keq
The reaction of hydrogen bromide(g) with chlorine(g) to form hydrogen chloride(g) and bromine(g) proceeds as follows: 2HBr(g) + Cl2(g)2HCl(g) + Br2(g) When 23.9 g HBr(g) reacts with sufficient Cl2(g), 12.0 kJ is evolved. Calculate the value of rH for the chemical equation given.
Answer:
The enthalpy of reaction per mole of HBr for this reaction = ΔrH = -40.62 kJ/mole.
Explanation:
2HBr(g) + Cl2(g) → 2HCl(g) + Br2(g)
When 23.9 g HBr(g) reacts with sufficient Cl2(g), 12.0 kJ of heat is evolved, calculate the value of ΔrH for the chemical reaction.
Note that ΔrH is the enthalpy per mole for the reaction.
Molar mass of HBr (g) = 80.91 g/mol.
Hence, 1 mole of HBr = 80.91 g
23.9 g of HBr led to the reaction giving off 12.0 kJ of heat
80.91 g of HBr will lead to the evolution of (80.91 × 12/23.9) = 40.62 kJ heat is given off.
Hence, 40.62 kJ of heat is given off per 80.91 g of HBr.
This directly translates to that 40.62 kJ of heat is given off per 1 mole of HBr
Hence, the heat given off per mole of HBr for this reaction is 40.62 kJ/mole.
But since the reaction liberates heat, it means the reaction is exothermic and the enthalpy change for the reaction (ΔHrxn) is negative.
Hence, ΔrH = -40.62 kJ/mole.
Hope this Helps!!!
For some hypothetical metal the equilibrium number of vacancies at 750°C is 2.8 × 1024 m−3. If the density and atomic weight of this metal are 5.60 g/cm3 and 65.6 g/mol, respectively, calculate the fraction of vacancies for this metal at 750°C.
Answer:
The correct answer is 5.447 × 10⁻⁵ vacancies per atom.
Explanation:
Based on the given question, the at 750 degree C the number of vacancies or Nv is 2.8 × 10²⁴ m⁻³. The density of the metal is 5.60 g/cm³ or 5.60 × 10⁶ g/m³. The atomic weight of the metal given is 65.6 gram per mole. In order to determine the fraction of vacancies, the formula to be used is,
Fv = Nv/N------ (i)
Here Nv is the number of vacancies and N is the number of atomic sites per unit volume. To find N, the formula to be used is,
N = NA×P/A, here NA is the Avogadro's number, which is equivalent to 6.022 × 10²³ atoms per mol, P is the density and A is the atomic weight. Now putting the values we get,
N = 6.022 × 10²³ atoms/mol × 5.60 × 10⁶ g/m³ / 65.6 g/mol
N = 5.14073 × 10²⁸ atoms/m³
Now putting the values of Nv and N in the equation (i) we get,
Fv = 2.8 × 10²⁴ m⁻³ / 5.14073 × 10²⁸ atoms/m^3
Fv = 5.44669 × 10⁻⁵ vacancies per atom or 5.447 × 10⁻⁵ vacancies/atom.
Clues about the history of the earth have been obtained from the study of
fossil fuels
rain forest materials
soll samples
O synthetic plastics
Answer: Fossil fuels
Explanation:
Fossil fuels such as petroleum, oil, and natural gas, are non-renewable energy resources which are formed from the remains of prehistoric ancient plants and animals beneath layers of rock of the earth surface.
By analyzing and studying fossil fuels using Radiocarbon analyses by archaeologists, earth scientists etc, Information about the history of the earth can be obtained from the decomposition of dead organisms present in fossil fuels.
From the unbalanced reaction: B2H6 + O2 ---> HBO2 + H2O
How many grams of O2 (32g/mol) will be needed to burn 36.1 g of B2H6 (Molar mass = 27.67g/mol)? ______g
Include the correct number of significant figures in your final answer
Answer: 125 g
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} B_2H_6=\frac{36.1g}{17}=1.30moles[/tex]
The balanced reaction is:
[tex]B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O[/tex]
According to stoichiometry :
1 mole of [tex]B_2H_6[/tex] require = 3 moles of [tex]O_2[/tex]
Thus 1.30 moles of [tex]B_2H_6[/tex] will require=[tex]\frac{3}{1}\times 1.30=3.90moles[/tex] of [tex]O_2[/tex]
Mass of [tex]O_2=moles\times {\text {Molar mass}}=3.90moles\times 32g/mol=125g[/tex]
Thus 125 g of [tex]O_2[/tex] will be needed to burn 36.1 g of [tex]B_2H_6[/tex]
Aqueous hydrobromic acid reacts with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . If of sodium bromide is produced from the reaction of of hydrobromic acid and of sodium hydroxide, calculate the percent yield of sodium bromide.
Answer:
The percentage yield is 50%
An organic chemistry student was studying the solubility of Methyl-N-acetyl-α-D-glucosaminide (1-O-methyl-GlcNAc), a derivative of glucosamine, in water but inadvertently added 1 equiv. of periodic acid instead. Based on your understanding of the reactions of monosaccharides with periodates, draw the organic product that the student obtained.
Complete Question
The diagram for this question is shown on the second uploaded image
Answer:
The organic product obtained is shown on the first uploaded image
Explanation:
The process that lead to this product formation is known as oxidative cleavage which is a reaction that involves the cleavage of a carbon to carbon bond at the same time this carbon which formed the carbon bond are oxidized i.e oxygen is been added to them
Solids in which the atoms have no particular order or pattern are called what solids
Answer:
Amorphous solids .
Explanation:
They have no particular order or pattern.Each particle is in a particular spot, but the particles are in no organized pattern.
Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 3.4 g of octane is mixed with 15.6 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits.
Answer:
10 g of CO2
Explanation:
Equation of the reaction:
CH3(CH2)6CH3 + 17O2 ----> 18H2O + 8CO2
Fom the above balanced equation,
1 mole of Octane gas reacts with 17 moles of oxygen gas to produce 8 moles of CO2
Molar mass of Octane = 114 g/mol
Molar mass of oxygen gas = 32 g/mol
Molar mass of CO2 = 44 g/mol
Therefore, 114 g of Octane reacts completely with 17 * 32g (= 544 g) of oxygen to produce 8 * 44 g(=352g) of CO2.
From the given mass of reactants;
3.4 g of Octane will react with (544 * 3.4)/114 g of oxygen = 16.22g of oxygen.
Therefore oxygen is the limiting reactant.
15.6 g of oxygen will react with (114 * 15.6)/544 g of CO2 = 3.27 g of octane.
Mass of CO2 produced will be
(352 * 15.6)/544 = 10 g of CO2
The breaking buffer that we use this week contains 10mM Tris, pH 8.0, 150mM NaCl. The elution buffer is breaking buffer that also contains 300mM imidazole. Describe how the instructor made the 0.25L elution buffer for all the students this week given 500ml of 1M of Tris (121.1 g/mole) (pH8.0), 750ml of 5M NaCl (MW
Answer:
Explanation:
From the given information ;the objective is to determine how the instructor made the 0.25L elution buffer
0.25 L elution buffer = 250 mL elution butter
The breaking buffer that we use this week contains
10mM Tris = 0.01 M
150mM NaCl = 0.15 M
300mM imidazole. = 0.3 M
The stock concentration of Tris in 1M
Therefore ; by using the formula: [tex]M_1V_1 = M_2 V_2[/tex]; we can determine the volume in the preparation; so;
[tex]1*V_1 = 0.0 1 \ M * 250 \ mL[/tex]
[tex]V_1 = \dfrac{0.0 1 \ M * 250 \ mL}{1 }[/tex]
[tex]V_1 = 2.5 \ mL[/tex]
In NaCl, The amount of stock concentration is 5 M
so; using the same formula; we have:
[tex]5*V_1 = 0.15 \ M * 250 \ mL[/tex]
[tex]V_1 = \dfrac{0. 15 \ M * 250 \ mL}{5 }[/tex]
[tex]V_1 = 7.5 \ mL[/tex]
From Imidazole ; the amount of stock concentration is
[tex]1*V_1 = 0.3 \ M * 250 \ mL[/tex]
[tex]V_1 = \dfrac{0. 3 \ M * 250 \ mL}{1 }[/tex]
[tex]V_1 = 75 \ mL[/tex]
Thus; we can have a table as shown as :
Stock concentration volume to be added Final concentration
1 M of Tris 2.5 mL 10 mM
5 M of NaCl 7.5 mL 150 mM
1 M of Imidazole 75 mL 300 mM
In conclusion. the addition of all the volume make up the 250 mL elution buffer that is equivalent to 0.25 L.
