There are 8.16 × 10-³ moles of CO2 gas at 100°C with a volume of 250 mL at 760 mm Hg.
HOW TO CALCULATE NUMBER OF MOLES:
The number of moles of a sample of gas can be calculated using the following formula:
PV = nRT
Where;
P = pressure of gas (atm)V = volume (L)n = number of moles (mol)R = gas law constant (0.0821 Latm/molK)T = temperature (K)According to this question;
P = 760mmHg = 1 atmT = 100°C = 100 + 273 = 373KV = 250mL = 0.250Ln = ?1 × 0.250 = n × 0.0821 × 373
0.250 = 30.62n
n = 0.250 ÷ 30.62
n = 8.16 × 10-³mol
Therefore, there are 8.16 × 10-³ moles of CO2 gas at 100°C with a volume of 250 mL at 760 mm Hg.
Learn more about number of moles at: https://brainly.com/question/4147359
How many moles of iron is equivalent to 4.45 x 10^22 atoms of iron
Answer:
0.074 moles
Explanation:
For every mole (of any element), there are 6.022 x 10^23 atoms.
There are 4.45 x 10^22 atoms of iron.
To find the moles we divide the number of atoms by Avogadro's number
4.45 x 10^22 / 6.022 x 10^23 = 0.0738957
Don't forget sig figs
Answer:
[tex]\boxed {\boxed {\sf 0.0739 \ mol \ Fe}}[/tex]
Explanation:
We are asked to convert a number of iron atoms to moles of iron.
We will use Avogadro's Number for this, which is 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. For this problem, the particles are atoms of iron. There are 6.022 ×10²³ atoms of iron in 1 mole of iron.
We will also use dimensional analysis to solve this problem. To do this, we use ratios. Set up a ratio using the underlined information.
[tex]\frac {6.022 \times 10^{23} \ atoms \ Fe} {1 \ mol \ Fe}[/tex]
Since we are converting 4.45 × 10²² atoms of iron to moles, we multiply the ratio by that value.
[tex]4.45 \ \times 10^{22} \ atoms \ Fe *\frac {6.022 \times 10^{23} \ atoms \ Fe} {1 \ mol \ Fe}[/tex]
Flip the ratio. The value is the same, but it allows us to cancel the units of atoms of iron.
[tex]4.45 \ \times 10^{22} \ atoms \ Fe *\frac {1 \ mol \ Fe}{6.022 \times 10^{23} \ atoms \ Fe}[/tex]
[tex]4.45 \ \times 10^{22} *\frac {1 \ mol \ Fe}{6.022 \times 10^{23}}[/tex]
Condense into 1 fraction.
[tex]\frac {4.45 \ \times 10^{22} }{6.022 \times 10^{23}} \ mol \ Fe[/tex]
[tex]0.07389571571 \ mol \ Fe[/tex]
The original measurement of atoms ( 4.45 × 10²²) has 3 significaint figures, so our answer must have the same. For the number we calculated, that is the ten-thousandths place. The 9 to the right of this place (0.07389571571) tells us to round the 8 up to a 9.
[tex]0.0739 \ mol \ Fe[/tex]
There are approximately 0.0739 moles of iron in 4.45 × 10²² atoms of iron.
Why do we need Chemistry in Nursing?
Answer:
We need chemistry in nursing because it deals with various kinds of drugs and the reactions of these drugs on the human body as well as with each other.
A solution has a higher boiling point than its associated pure solvent does.
What is this property of the solution called?
1. boiling-point depression
2. freezing-point depression
3. vapor-pressure lowering
4. boiling-point elevation
Answer:
4 boiling point elevation
A capsule containing 0.500 L of air at 1.00 atm is compressed to 3.25 atm. At that point, what is the volume of the gas in the capsule?
Answer:
V₂ = 0.154 Liters
Explanation:
Pressure => P
Volume => V
Temperature => T
mass (moles) => n
This problem...
P₁ = 1.00 ATM P₂ = 3.25 ATM
V₁ = 0.500L V₂ = ?
T₁ = constant T₂ = T₁ = constant
n₁ = constant n₂= n₁ = constant
P₁V₁/n₁T₁ = P₂V₂/n₂T₂ => V₂ = V₁(P₁/P₂) = 0.500L (1.00ATM/3.25ATM) = 0.154 Liters
how many moles of KF are present in 46.5 grams of KF
Explanation:
here's the answer to your question
Answer:
0.8017
Explanation:
Find the molar Mass of KF
K = 39
F = 19
Total = 58
Note: these numbers are approximate. Use your periodic table to get the exact numbers.
mols = given mass / molar mass
given mass = 46.5
molar mass = 58
mols = 46.5 / 58
mols = 0.8017
WHAT WOULD THE RIGHT OPTION??
how many resonance structures of benzene are known?
A) 3
B) 4
C) 5
D) 6
Answer:
C) 5
Step-by-step explaination:
Benzene has 5 resonance structures.
Starting from (R)-3-methylhex-1-yne as the substrate at the center of your page, draw a reaction map showing the regiochemical and stereochemical outcome or outcomes for each of the following series of reagents. Name each of your products, including stereochemical designations for any chirality centers that are generated.
a. HgSO4, H2SO4, H2O
b. 1. 9-BBN; 2. H2O2, NaOH
c. Br2, CCl4
d. HBr
Solution :
A substrate is defined as the chemical species that are being observed in the chemical reaction where the substrate reacts with a reagent and forms a product. It can also be referred to the surface where some other chemical reactions are performed.
Stereochemistry is defined as the study of relative spatial arrangement of the atoms which forms the structure of the molecules and their respective manipulations.
In the context, the products including the stereochemical designations for any chirality centers starting from the (R)-3-methylhex-1-yne as the substrate are attached below.
Phosphorus-32 is radioactive and has a half life of 14.3 days. Calculate the activity of a 3.5mg sample of phosphorus-32. Give your answer in becquerels and in curies. Round your answer to 2 significant digits.
Answer:
The activity of P-32 is 3.7x10¹³ becquerels = 1.0x10³ curies.
Explanation:
The activity of P-32 can be calculated with the following equation:
[tex] A = \lambda N [/tex] (1)
Where:
N: is the number of atoms of P-32
λ: is the decay constant
We can find the number of atoms of P-32 as follows:
[tex] N = \frac{N_{A}*m}{M} [/tex] (2)
Where:
[tex]N_{A}[/tex]: is the Avogadro's number = 6.022x10²³ atoms/mol
m: is the mass of P-32 = 3.5x10⁻³ g
M: is the molar mass of the radionuclide (P-32) = 32 g/mol
Now, the decay constant is given by:
[tex] \lambda = \frac{ln(2)}{t_{1/2}} [/tex] (3)
Where:
[tex]{t_{1/2}} [/tex]: is the half-life of P-32 = 14.3 days
Finally, we can find the activity of P-32 by entering equations (2) and (3) into (1):
[tex] A = \lambda N = \frac{ln(2)}{t_{1/2}}*\frac{N_{A}*m}{M} = \frac{ln(2)}{14.3 d*\frac{24 h}{1 d}*\frac{3600 s}{1 h}}*\frac{6.022 \cdot 10^{23} mol^{-1}*3.5 \cdot 10^{-3} g}{32 g/mol} = 3.7 \cdot 10^{13} dis/s [/tex]
Since a becquerel (Bq) is defined as a disintegration (dis) per second, the activity in Bq is:
[tex] A = 3.7 \cdot 10^{13} Bq [/tex]
And, since a Curie (Ci) is 3.7x10¹⁰ Bq, the activity in Ci is:
[tex] A = 3.7 \cdot 10^{13} Bq*\frac{1 Ci}{3.7 \cdot 10^{10} Bq} = 1.0 \cdot 10^{3} Ci [/tex]
Therefore, the activity of P-32 is 3.7x10¹³ becquerels = 1.0x10³ curies.
I hope it helps you!
Identify “A” in the following reaction: CH3¬COOH + Na2CO3 → A + CO2 + H2O
The windscreen of a window is made up of
Answer:
Modern windshields are generally made of laminated safety glass, a type of treated glass, which consists of, typically, two curved sheets of glass with a plastic layer laminated between them for safety, and bonded into the window frame.
Explanation:
mark me brain liest if my answer is correct
When (R)-2-chloro-3-methylbutane is treated with potassium tert-butoxide, a monosubstituted alkene is obtained. When this alkene is treated with HBr, a mixture of products is obtained. Identify all of the expected products.
Answer:
See explanation and image attached
Explanation:
The reaction of (R)-2-chloro-3-methylbutane with potassium tert-butoxide yields a monosubstituted alkene .
Since the base is bulky, the Hoffman product predominates because attack occurs at the less hindered carbon atom to yield the major product as shown.
The alkene reacts with HBr at the secondary carbon atom to yield a carbocation intermediate which is flat and planar. Attack on either face of the carbocation yields a racemic mixture of the (2R) and (2S) products.
Rearrangement of the carbocation to yield a tertiary carbocation gives the 2-bromo-2-methyl butane product as shown in the image attached.
What is the pressure in a 19.1 L cylinder filled with 0.684 mol of nitrogen gas at a temperature of 326 K ?
Answer:
The hydrogen sample has a pressure of 0.957 atmospheres.
Explanation:
Let consider that the hydrogen sample behaves ideally, the equation of state for ideal gases is:
[tex]P\cdot V = n\cdot R_{u}\cdot T[/tex] (1)
Where:
[tex]P[/tex] - Pressure, in atmospheres.
[tex]V[/tex] - Volume, in liters.
[tex]n[/tex] - Molar quantity, in moles.
[tex]T[/tex] - Temperature, in Kelvin.
[tex]R_{u}[/tex] - Ideal gas constant, in atmosphere-liters per mole-Kelvin.
If we know that [tex]V = 19.1\,L[/tex], [tex]n = 0.684\,mol[/tex], [tex]T = 326\,K[/tex] and [tex]R_{u} = 0.082\,\frac{atm\cdot L}{mol\cdot K}[/tex], then the pressure of the hydrogen sample is:
[tex]P = \frac{n\cdot R_{u}\cdot T}{V}[/tex]
[tex]P = \frac{(0.684\,mol)\cdot \left(0.082\,\frac{atm\cdot L}{mol\cdot K} \right)\cdot (326\,K)}{19.1\,L}[/tex]
[tex]P = 0.957\,atm[/tex]
The hydrogen sample has a pressure of 0.957 atmospheres.
Consider the molecule PF5.
Indicate how many lone pairs you would find on the central atom:
Indicate how many total bonds are connected to the central atom (count single bonds as 1 bond, double bonds as 2 bonds, and triple bonds as 3 bonds):
Explanation:
here's the answer to your question
What is the correct ratio of carbon to hydrogen to oxygen in glucose (C6H12O6)?
12:12:6
2:1:1
1:2:1
6:6:12
Answer:
Correct ratio of carbon to hydrogen is 2:1:1
Answer:
Its actually 1:2:1
Explanation:
The molecular formula is C6H12O6 because one molecule actually contains 6 C, 12 H, and 6 O atoms. The simplest whole-number ratio of C to H to O atoms in glucose is 1:2:1, so the empirical formula is CH2O.
HBr can be added to an alkene in the presence of peroxides (ROOR). What function does the peroxide serve in this reaction
Answer:
Radical chain initiator
Explanation:
The peroxide here serves as a radical chain initiator. In the field of chemistry the radical initiatives are those substances that are used in industrial processes like polymer synthesis. These initiatives have weak bonds generally and they're mostly used to create free radicals. These radicals are atoms that have odd numbers of electrons. Peroxide is an example of such.
Based on periodic properties, choose the more metallic element from each of the following pairs.
Match the words in the left column to the appropriate blanks in the sentences on the right.
Between Sr and Sb, the more metallic element is ______
Between \rm Sr and \rm Sb, the more metallic element is _______
Between As and Bi, the more metallic element is ______
Between \rm As and \rm Bi, the more metallic element is _______
Between Cl and O, the more metallic element is ______
Between \rm Cl and \rm O, the more metallic element is ______
Between S and As, the more metallic element is ______
Between \rm S and \rm As, the more metallic element is _______
Answer:
Sr is the more metallic element
Bi is the more metallic element
O is the more metallic element
As is the more metallic element
Explanation:
One thing should be clear; metallic character increases down the group but decreases across the period.
Hence, as we move across the period, elements become less metallic. As we move down the group elements become more metallic.
This is the basis upon which decisions were made about the metallic character of each of the elements listed above.
What is the balanced equation for the reaction of lithium metal with fluorine gas? Li ( s ) + F ( g ) → LiF ( s ) Li ( s ) + F 2 ( g ) → LiF ( s ) 2 Li ( s ) + F 2 ( g ) → 2 LiF ( s ) Li ( s ) + F 2 ( g ) → LiF
Answer:
2Li(s) + F2(g)→2LiF(s)
Explanation:
What is [H] for the solution?
x 100 M
n=
Answer:
Asumiendo
"M"
es una variable
|
Usar como
un número romano
en lugar de
Suponiendo la multiplicación
|
Uso una lista en lugar de
H x×100 M n
Figura geométrica
línea
Propiedad como función
Paridad
aun
Derivado
d/dx(H x×100 M n) = 100 H M n
Integral indefinida
integral100 H M n x dx = 50 H M n x^2 + constante
Integral definida sobre una hiperesfera de radio R
integral integral integral_(H^2 + M^2 + n^2 + x^2<R^2) 100 H M n x dH dM dn dx = 0
Integral definida sobre un hipercubo de longitud de borde 2 L
integral_(-L)^L integral_(-L)^L integral_(-L)^L integral_(-L)^L 100 H M n x dx dn dM dH = 0
Explanation:
Calculate the molality of each of the following solutions: (a) 36.2 g of sucrose (C12H22O11) in 323 g of water, m (b) 8.63 moles of ethylene glycol (C2H6O2) in 1889 g of water.
Answer:
(a) m = 0.327 m.
(b) m = 4.57 m.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by firstly considering the fact that the molality is computed by dividing the moles of solute by the kilograms of solvent, in this case water; in such a way, we proceed as follows:
(a) We firstly calculate the moles of 36.2 grams of sucrose as its molar mass is 342.3 g/mol:
[tex]\frac{36.2g}{342.3g/mol} =0.106mol[/tex]
Next, the kilograms of water in this case are 0.323 kg so that the molality will be:
[tex]m=\frac{0.106mol}{0.323kg}\\\\m=0.327m[/tex]
(b) In this case, we directly realize that the kilograms of water are now 1.889 kg so that the molality will be:
[tex]m=\frac{8.63mol}{1.889kg}=4.57m[/tex]
Clearly, the both of them in molal, m, units.
Regards!
Select the choice that best completes the following sentence: When cooled slowly, transformations near the melting temperature tend to yield ______ grains due to the formation of ______ nucleation sites followed by ______ grain growth.
Question Completion with Options:
O coarse...few...rapid
O fine...few...slow
O fine...multiple...rapid
O coarse...few...slow
O fine...multiple...slow
Answer:
The choice that best completes the sentence is:
O coarse...few...slow
Explanation:
Transformations near the melting temperature develop coarse grains because few nucleation sites are formed and the rate of the grain growth is usually slow. This is because of the process that starts with recrystallization, recovery, and nucleation before growth can occur. While recrystallization enables the grain to increase in size at high temperature, nucleation gives the grain the energy to irreversibly grow into larger-sized nucleus.
What is this organic compound?
Please asap!!
3,3-dimethylhexane is the nomenclature of the compound
Consider the reaction below to answer the following questions (4) a. The nucleophile in the reaction is _______ b. The Lewis acid catalyst in the reaction is ______ c. This reaction proceeds___________(faster or slower)
The question is incomplete, the complete question is shown in the image attached to this answer.
Answer:
a) Br^-
b) FeCl3
c) slower
d) see the first attached image
Explanation:
Aromatic compounds undergo electrophilic substitution sections in the presence of the appropriate electrophile.
In the reaction above, the Br^- nucleophile attacks the Lewis acid FeCl3. Recall that the nitro group is meta directing hence the incoming Br^+ electrophile is directed towards the meta position as shown in the image attached.
Note that the nitro group deactivates the ring towards electrophilic substitution hence the reaction is slower with nitrobenzene than with unsubstituted benzene.
Part A
When the following liquids are poured into the same container, they separate as shown in the image. Based on the data
in the table below, what caused the order of the layers?
rubbing alcohol
vegetable oil
water
corn syrup
Mass
Liquid
corn syrup
water
Volume Used
95 cm
90 cm
85 cm
105 cm?
130.158
90.00 8
77.358
81.908
Density
1.37 g/cm
1 g/cm
0.91 g/cm
0.78 g/cm
vegetable oil
rubbing alcohol
I
B
X
Font Sizes
A- A -
E 3
Answer: The layers are ordered by density, with the least dense layer on top, and the densest layer on the bottom.
Explanation:
Plato
What mass of oxygen is needed for the complete combustion of 1.60-10^-3
g
of methane?
Express your answer with the appropriate units.
Answer:
6.4×10¯³ g of O₂.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CH₄ + 2O₂ —> CO₂ + 2H₂O
Next, we shall determine the masses of CH₄ and O₂ that reacted from the balanced equation. This can be obtained as follow:
Molar mass of CH₄ = 12 + (4×1)
= 12 + 4
= 16 g/mol
Mass of CH₄ from the balanced equation = 1 × 16 = 16 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mass of O₂ from the balanced equation = 2 × 32 = 64 g
SUMMARY:
From the balanced equation above,
16 g of CH₄ reacted with 64 g of O₂.
Finally, we shall determine the mass of O₂ needed to react with 1.6×10¯³ g of CH₄. This can be obtained as illustrated below:
From the balanced equation above,
16 g of CH₄ reacted with 64 g of O₂.
Therefore, 1.6×10¯³ g of CH₄ will react with = (1.6×10¯³ × 64) / 16 = 6.4×10¯³ g of O₂
Thus, 6.4×10¯³ g of O₂ is needed for the reaction.
4. A sample of ammonia, NH3, contains 3.3 x 1021 hydrogen atoms. How many NH; molecules are in this sample?
Answer:
1.1 × 10²¹ NH₃ molecules
Explanation:
From the given information:
We were being told that the number of the hydrogen (H) atoms present in the sample of NH3 = 3.3 × 10²¹ hydrogen.
However, it signifies that each molecule of ammonia harbors 3hydrogen (H) atoms.
Hence, the number of molecules of NH₃ present;
[tex]\mathsf{=\dfrac{3.3\times 10^{21}}{3} \ molecules \ of \ {NH_3}}[/tex]
= 1.1 × 10²¹ NH₃ molecules
Do you think that people decide their destiny, or that their destiny has already been decided and they cannot change ,they only discover it ?
i don't know if this is for school or what. but I believe that your destiny is set out but you do have the power to change it. you decide what path you want to go down with your actions
An enzyme acts to Group of answer choices raise the activation energy needed to start the reaction. lower the activation energy needed to start the reaction. convert the activation energy into potential energy. convert the activation energy into kinetic energy. stop a chemical reaction.
Answer:
lower the activation energy needed to start the reaction.
Explanation:
The activation energy is defined as the energy barrier that stands between reactants and products.
An enzyme is a biological catalyst. Catalysts are known to lower the activation energy of a reaction.
Hence, a catalyst lowers the activation energy of the reaction. The lower the activation energy of a reaction, the faster the reaction is expected to be.
According to the ideal gas law, a 9.998 mol sample of argon gas in a 0.8311 L container at 502.7 K should exert a pressure of 496.2
atm. What is the percent difference between the pressure calculated using the van der Waals' equation and the ideal pressure? For Ar
gas, a = 1.345 L’atm/mol? and b = 3.219x10-2 L/mol.
Pideal – Puan der Waals |
Percent difference
x 100
Answer:
[tex]\%diff=24.0\%[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to set up the van der Waals' equation as shown below:
[tex]p=\frac{RT}{v-b}-\frac{a}{v^2}[/tex]
Thus, we secondly calculate the molar volume as:
[tex]v=\frac{0.8311L}{9.998mol} =0.083L/mol[/tex]
Then, we plug in the entire variables in the vdW equation to get such pressure:
[tex]p=\frac{0.08206\frac{atm*L}{mol*K}*502.7K}{0.08313L/mol-0.03219L/mol}-\frac{1.345L*atm/mol}{(0.08313L/mol)^2}\\\\p=615.2atm[/tex]
And the ideal gas pressure:
[tex]p=\frac{0.08206\frac{atm*L}{mol*K}*502.7K}{0.08313L/mol}\\\\p=496.2atm[/tex]
Finally, the percent difference:
[tex]\%diff=\frac{|496.2atm-615.2atm|}{496.2atm} *100\%\\\\\%diff=24.0\%[/tex]
Regards!
Titanium is a metal often used as an alloying agent to provide materials that are strong, lightweight, and temperature-resistant Which of the following represents the correct ground-state configuration for a neutral atom of titanium?
A) 1s 2s 2p 3s 3p 48°30°
B) 1s 2s 2p 3s 3p 4s3d
C) 15*2s2p 3s 3p 4s
D) 15°2s 2p 3s 3p 3d
Answer:B) 1s 2s 2p 3s 3p 4s 3d
Explanation:
The ground state electron configuration shows how the electrons in the atomic orbitals of an atom are in their lowest , most stable energy arrangements and since Electrons must be filled following the Aufbau's principle(electrons fill lowest energy shells first)
Now, Titanium lies in period IV and group 4 of the periodic table with 22 as its atomic number
Thus, the ground-state electron configuration of a neutral atom of titanium is 1s²2s²2p⁶3s²3p⁶4s²3d².
What volume of carbon dioxide is required for inflating the Ziploc bag prototype ?
Answer:
The front passenger airbag has a volume of around 140 l and fully inflates in around 35 ms. The process is similar for side airbags (thorax airbags).
