Which phrase is evidence of the limits of what science can study?
This question is incomplete; here is the complete question:
Which phrase is evidence of the limits of what science can study?
A. Only a phenomenon that a scientist can observe with instruments
B. Only a phenomenon that a scientist can observe without instruments
C. Only a phenomenon that a scientist will be able to explain with certainty
D. Any phenomenon that a scientist can observe or model
The correct answer is D. Any phenomenon that a scientist can observe or model
Explanation:
Science focuses on phenomena that can be objectively studied or witnessed. This includes any phenomenon a scientist can register using observation, instruments, or that can be represented through models. This is because, through observations or instruments, it is possible to prove the phenomenon occurs and understand how this happens; also, models can represent or recreate the phenomenon itself, which is essential to understand and explain the phenomenon. Moreover, science has an objective focus and due to this, aspects such as personal beliefs that cannot be represented through a model or be observed are not part of science. According to this, the correct answer is D.
Why do astronauts with less on the moon than they do on Earth
The Ka values for several weak acids are given below. Which acid (and its conjugate base) would be the best buffer at pH 3.7?a. MES: Ka 7.9 x 10b. HEPES; Ka 3.2 x 103c. Tris; Ka 6.3 x 109d. Formic acid: K 1.8 x 10
e. Acetic acid: K 1.8 x 10
Answer:
Formic acid and Acetic acid is the best buffer at pH 3.7.
Explanation:
Given that,
The Ka values for several weak acids are given,
[tex]K_{a}\ of\ MES=7.9\times10^{-7}[/tex]
[tex]K_{a}\ of\ HEPES = 3.2\times10^{-3}[/tex]
[tex]K_{a}\ of\ Tris=6.3\times10^{-9}[/tex]
[tex]K_{a}\ of\ formic\ acid = 1.8\times10^{-4}[/tex]
[tex]K_{a}\ of\ Acetic\ acid = 1.8\times10^{-5}[/tex]
We need to calculate the pH of the weak acids with their [tex]pK_{a}[/tex] values
Using formula of [tex]pK_{a}[/tex]
For MES,
[tex]pK_{a}=-log K_{a}[/tex]
Put the value into the formula
[tex]pK_{a}=-log(7.9\times10^{-7})[/tex]
[tex]pK_{a}=7.0-log7.9[/tex]
[tex]pK_{a}=6.1[/tex]
pH range for best buffer,
[tex]pH=pK_{a}\pm 1[/tex]
Put the value into the formula
[tex]pH=6.1\pm 1[/tex]
[tex]pH=7.1, 5.1[/tex]
The pH value of the solution between 7.1 to 5.1.
This is not best buffer.
For HEPES,
[tex]pK_{a}=-log K_{a}[/tex]
Put the value into the formula
[tex]pK_{a}=-log(3.2\times10^{-3})[/tex]
[tex]pK_{a}=3.0-log3.2[/tex]
[tex]pK_{a}=2.5[/tex]
pH range for best buffer,
[tex]pH=pK_{a}\pm 1[/tex]
Put the value into the formula
[tex]pH=2.5\pm 1[/tex]
[tex]pH=3.5, 1.5[/tex]
The pH value of the solution between 3.5 to 1.5.
This is not best buffer.
For Tris,
[tex]pK_{a}=-log K_{a}[/tex]
Put the value into the formula
[tex]pK_{a}=-log(6.3\times10^{-9})[/tex]
[tex]pK_{a}=9.0-log6.3[/tex]
[tex]pK_{a}=8.2[/tex]
pH range for best buffer,
[tex]pH=pK_{a}\pm 1[/tex]
Put the value into the formula
[tex]pH=8.2\pm 1[/tex]
[tex]pH=9.2, 7.2[/tex]
The pH value of the solution between 9.2 to 7.2.
This is not best buffer.
For formic acid,
[tex]pK_{a}=-log K_{a}[/tex]
Put the value into the formula
[tex]pK_{a}=-log(1.8\times10^{-4})[/tex]
[tex]pK_{a}=4.0-log1.8[/tex]
[tex]pK_{a}=3.7[/tex]
pH range for best buffer,
[tex]pH=pK_{a}\pm 1[/tex]
Put the value into the formula
[tex]pH=3.7\pm 1[/tex]
[tex]pH=4.7, 2.7[/tex]
The pH value of the solution between 4.7 to 2.7.
This is best buffer.
For acetic acid,
[tex]pK_{a}=-log K_{a}[/tex]
Put the value into the formula
[tex]pK_{a}=-log(1.8\times10^{-5})[/tex]
[tex]pK_{a}=5.0-log1.8[/tex]
[tex]pK_{a}=4.7[/tex]
pH range for best buffer,
[tex]pH=pK_{a}\pm 1[/tex]
Put the value into the formula
[tex]pH=4.7\pm 1[/tex]
[tex]pH=5.7, 3.7[/tex]
The pH value of the solution between 5.7 to 3.7.
This is best buffer
Hence, Formic acid and Acetic acid is the best buffer at pH 3.7.
What is the density of an unknown metal that has a mass of 157.4 g and a volume of 20 mL?
Answer:
7.87g/mL
Explanation:
density=mass/volume
density=157.4/20
density=7.87g/mL
unit processes in charge of smelting
Answer:
Smelting, process by which a metal is obtained, either as the element or as a simple compound, from its ore by heating beyond the melting point, ordinarily in the presence of oxidizing agents, such as air, or reducing agents, such as coke.
Hope I helped!
Explanation:
general characteristics of coinage metals
Answer:
Characteristics. They are all relatively inert, corrosion-resistant metals. Copper and gold are colored. These elements have low electrical resistivity so they are used for wiring.
Explanation:
Hope this help u
️️
Find the density of an empty tissue box with a mass of 345 grams and a volume of 1125 cm3.
Answer:
The answer is
0.307 g/cm³Explanation:
The density of a substance can be found by using the formula
[tex]density = \frac{mass}{volume} [/tex]
From the question
mass of box is 345 g
volume = 1125 cm³
The density of the object is
[tex]density = \frac{345}{1125} \\ = 0.3066666...[/tex]
We have the final answer as
0.307 g/cm³Hope this helps you
help me ill give brainliest i need help with a b c d and to see if i got propene and methane right please help :(
Answer:Yes, you got it right
Explanation:
Pls give brainliest
An automobile tire contains air at 320.×103 Pa at 20.0 ◦C. The stem valve is removed and the air is allowed to expand adiabatically against the constant external pressure of 100.×103 Pa until P = Pexternal. Assume the air is an ideal gas with C¯ V = 5/2 R (diatomic). Calculate the final temperature.
Answer:
6.15.3 k
Explanation:
From the question we can see that
q = 0, Δu = w
Then,
[tex]T_f = \frac{C_{V,m}+RP_{ext}P_i}{C_{V,m}+RP_{ext}P_f} T_i[/tex]
putting values wet
=[tex]\frac{2.5\times 8.314+8.314\left(10^5\right)\left(3.20\times 10^5\right)}{2.5\times 8.314+\left(8.314\right)\left(10^5\right)\left(10^5\right)}\times \:293[/tex]
T_f = 615.3 K
What is most likely to occur when jagged edges of rock plates grind past each other?
Answer:
The most likely to occur when jagged edges of rock plates grind past each other is the presence of a high degree of frictional force.
This may cause the rocks to be broken down into smaller particles.
It also implies that the energy necessary for further disintegration and movement of rocks is stored up.
the simplest formula for a compound made from element X(molar mass=79.0g mol) that is 21.0% nitrogen by mass is ___
A. XN
B. XN2
C.X2N2
D.X3N2
Answer:
C
Explanation:
Compounds are substances that are made of elements linked by chemical bonds. "The simplest formula for a compound made from element X that is 21.0% nitrogen by mass is X₂N₃." Thus, option C is correct.
What is the empirical formula?The empirical formula has been the representation of the ratio of the whole number of atoms involved to make a molecule or compound. It has been calculated by calculating the moles through mass and, molar mass.
The elements involved are X (unknown), and N (known).
Given,
Mass percentage of nitrogen = 21.0% = 21 gms
Mass percentage of X = 100.0 % - 21.0 % = 79.0% = 79 gms
Molar mass of X = 79.0 g /mol
Molar mass of Nitrogen = 14 g/mol
Moles of X is calculated as:
Moles X = 79 ÷ 79
= 1 mol
Moles of N are calculated as:
Moles N = 21 ÷ 14
= 1.5 mol
Now, the moles are divided by the smallest mole as:
X = 1 ÷ 1 = 1
N = 1.5 ÷ 1 = 1.5
The ratios are 1:1.5 or 2:3. 2X and 3N.
The empirical formula for the compound will be X₂N₃.
Therefore, the simplest formula for the compound will be X₂N₃.
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why do scientists store oxygen and hydrogen in water?
Answer:
When molecular hydrogen (H2) and oxygen (O2) are combined and allowed to react together, energy is released and the molecules of hydrogen and oxygen can combine to form either water or hydrogen peroxide. ... For both of the reactions shown, the hydrogen molecules are oxidized and the oxygen atoms are reduced.
Explanation:
Select the numbers that show how the numbers given in scientific notation can be added. 2.00 times 10 to the negative 26th power minus 3.34 time 10 to the negative 27th power. Below that, a math example: 2.00 x 10 to the negative 26th power minus A time 10 to the B power equals C times 10 to the D power. A = B = C = D =
Answer:
A = 0.334
B = -26
C = 1.666
D = -26
Explanation:
just took quiz
Answer:
A = ✔ 0.334
B = ✔ -26
C = ✔ 1.666
D = ✔ -26
Explanation:
edge 2023
My barometer is reading 1253 torr. What is the pressure in mmHg?
A. 1.253
B. 1.649
C. 1253
D. 1649
Answer:
The pressure in mmHg is 1253 (option C)
Explanation:
Two quantities are directly proportional if when multiplying or dividing one of them by a number, the other is multiplied or divided by that number. In other words, the magnitudes are directly proportional when one magnitude increases and so does the other in the same proportion; or when one magnitude decreases and so does the other in the same proportion.
The rule of three or is a way of solving proportionality problems between three known values and an unknown value, which can be applied to directly proportional quantities as follows:
a ⇒ b
c ⇒ x
So [tex]x=\frac{c*b}{a}[/tex]
where a, b and c are data and x is the unknown value to be calculated.
In this case, knowing that 1 Torr = 1 mmHg, the rule of three can be applied as follows: if 1 torr is equal to 1 mmHg, 1253 torr is equal to how many mmHg?
[tex]pressure=\frac{1253 torr*1 mmHg}{1 torr}[/tex]
pressure= 1253 mmHg
The pressure in mmHg is 1253 (option C)
[H+] for a solution is 4.59 x 10-6 M.
This solution is
A. acidic
B. basic
C. neutral
Answer:
A. acidic
Explanation:
Answer:
acidic
Explanation:
If the atom has a MASS of 127 and 45 NEUTRONS, then it contains how many
PROTONS

Answer:
53
Explanation:
because minus 3
Which statements are true about triangle ABC and it’s translated image, ABC? Select two options.
HURRY PLEASE! True or false: Making an observation is the first step of the scientific method.
Answer:
True
Explanation:
You must first observe your data then form a hypothesis.
The diffusion coefficient for aluminum in silicon is DAl in Si= 4 × 10-13 cm2/s at 1300 K. What is a reasonable value for DAl in Si at 1600 K ? Note: Rather than performing a specific calculation, you should be able to justify your answer from the options below based on the mathematical temperature dependence of the diffusion coefficient assuming a positive activation energy for diffusion.
Answer:
D = 4x10^-11
Explanation:
an increase in temperature would cause a resultant increase in diffusivity. as temperature rises, thermal energy of atoms would also rise and this would cause them to go faster.
4x10^-11cm²/s satisfies this condition because there is a temperature increase from 1300 to 1600.
DAI in Si = 4x10^-13cm²/sec at 1300
DAI in Si at 1600
D increases
temperature also increases
Argon (Ar) and helium (He) are initially in separate compartments of a container at 25°C. The
Ar in compartment A, which has a volume VA of 6.00 L, has a pressure of 2.00 bar. The He in
compartment B of unknown volume V3 has a pressure of 5.00 bar. When the two compartments
are connected and the gases allowed to mix, the total pressure of gas is 3.60 bar. Assume both
gases behave ideally
(a) [4 marks) Determine the volume of compartment B.
(b) [2 marks] Determine the mole fraction of He in the mixture of gases.
Answer:
(a) [tex]V_B=11.68L[/tex]
(b) [tex]x_{He}=0.533[/tex]
Explanation:
Hello,
In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:
[tex]n_A=\frac{0.082\frac{atm*L}{mol*K}*298K}{1.974 atm*6.00L}=2.063mol[/tex]
Thus, since the final pressure is 3.60 bar, we can write:
[tex]P=x_{Ar}P_A+x_{He}P_B\\\\P=\frac{n_{Ar}}{n_{Ar}+n_{He}} P_A+\frac{n_{He}}{n_{Ar}+n_{He}} P_B\\\\3.60bar=\frac{2.063mol}{2.063mol+n_{He}} *2.00bar+\frac{n_{He}}{2.063mol+n_{He}} *5.00bar[/tex]
The moles of helium could be computed via solver as:
[tex]n_{He}=2.358mol[/tex]
Or algebraically:
[tex]3.60bar=\frac{1}{2.063mol+n_{He}} *(4.0126+5.00*n_{He})\\\\7.314+3.60n_{He}=4.013+5.00*n_{He}\\\\7.314-4.013=5.00*n_{He}-3.60n_{He}\\\\n_{He}=\frac{3.3}{1.4}=2.358mol[/tex]
In such a way, the volume of the compartment B is:
[tex]V_B=\frac{n_{He}RT}{P_B}=\frac{2.358mol*0.082\frac{atm*L}{mol*K}*298.15K}{4.935atm}\\ \\V_B=11.68L[/tex]
Finally, he mole fraction of He is:
[tex]x_{He}=\frac{2.358}{2.358+2.063}\\ \\x_{He}=0.533[/tex]
Regards.
A beaker contains 9.80 L
of water. What is the
volume in quarts
Answer:
10.4 qt
Explanation:
Step 1: Given data
Volume of water in the beaker (V): 9.80 L
Step 2: Convert the volume of water in the beaker to US quarts (qt)
In order to convert one unit into another, we need a conversion factor. In this case, the appropriate conversion factor is 1 L = 1.06 qt. The volume of water in the beaker, in US quarts, is:
9.80 L × (1.06 qt/1 L) = 10.4 qt
. Answer the following: 7 x 2 = 14
1. Find
1. Find
2/3÷4
Explanation:
2/3 ÷ 4 = 0.166
I think this should be the answer
Explanation:
[tex] \frac{2}{3} \div 4[/tex]
[tex]make \: 4 \: as \: \frac{1}{4} \: since \: is \: divide[/tex]
[tex] \frac{2}{3} \times \frac{1}{4} [/tex]
[tex] \frac{2}{12} = \frac{1}{6} [/tex]
Hope this is correct and helpful
HAVE A GOOD DAY!
Determine which physical conditions are necessary to support nuclear fusion and formation of stars.
Answer:
The correct approach will be "Increased gravitational attraction".
Explanation:
The increased gravitational attraction seems to be the natural or physical phenomenon that is required to promote nuclear reactions including star formation. This is much more important to establish stars at different temperatures (lower) and greater magnetic pull as well as nuclear fusion tends to happen.Strontium-90 is radioactive and has a half life of 28.8 years. How much of a 2.90mg sample would be left after 137. years?
Round your answer to 2 significant digits. Also, be sure your answer has a unit symbol.
Answer:
0.11 mg
Explanation:
From the question given above, the following were obtained:
Half life (t½) = 28.8 years
Original amount ((N₀) = 2.90 mg
Time (t) = 137 years
Amount remaining (N) =?
Next, we shall determine the rate of disintegration (K). This can be obtained as follow:
Rate of decay (K) = 0.693/ half life (t½)
K = 0.693 / t½
Half life (t½) = 28.8 years
Rate of decay (K) =?
K = 0.693 / t½
K = 0.693 / 28.8
K = 0.0241 /year
Finally, we shall determine amount remaining after 137 years as follow:
Original amount ((N₀) = 2.90 mg
Time (t) = 137 years
Rate of decay (K) = 0.0241 /year
Amount remaining (N) =?
Log (N₀/N) = kt/2.3
Log (2.9/N) = 0.0241 × 137 / 2.3
Log (2.9/N) = 3.3017 / 2.3
Log (2.9/N) = 1.4355
Take the antilog of 1.4355
2.9/N = antilog (1.4355)
2.9/N = 27.26
Cross multiply
2.9 = N × 27.26
Divide both side by 27.26
N = 2.9/27.26
N = 0.11 mg
Therefore, the amount remaining after 137 years is 0.11 mg
what is the molar mass of PbCO₄
Answer:
283.211
Explanation:
Which of the following species is the least nucleophilic?A) (CH3)3NB) BF3C) H2OD) (CH3)3CO-E) CN-
Answer:
BF3
Explanation:
BF3 possesses an open sextet. This implies that the octet around the central atom is incomplete. As a result of this, BF3 will function as an electrophile rather than as a nucleophile.
Hence of all the compounds listed in the options, BF3 is the least nucleophillic.
Consider the fructose-1,6-bisphosphatase reaction. Calculate the free energy change if the ratio of the concentrations of the products to the concentrations of the reactants is 21.321.3 and the temperature is 37.0°C37.0°C ? ΔG°′ΔG°′ for the reaction is −16.7 kJ/mol−16.7 kJ/mol .
Answer:
ΔG = -8.812 kJ/mol
Explanation:
To obtain the free energy of a reaction you can use the expression:
ΔG = ΔG° + RT ln Q
Where:
ΔG° is Standard Gibbs Free energy: -16.7kJ/mol = -16700J/mol
R is gas constant: 8.314472 J/molK
T is absolute temperature (37°C + 273.15 = 310.15K)
And Q is reaction quotient: 21.3
Replacing in the formula:
ΔG = ΔG° + RT ln Q
ΔG = -16700J/mol + 8.314472J/molK*310.15K ln 21.3
ΔG = -8812.4J/mol
ΔG = -8.812 kJ/mol
Which one correctly lists the compound that has the highest boiling point from each group it belongs to? Group I: SiH 4, SnH 4, CH 4, and GeH 4 Group II: HCl, HBr, HF, and HI Group I: SiH4 and Group II: HF Group I: SnH4 and Group II: HF Group I: SnH4 and Group II: HI Group I: CH4 and Group II: HI Group I: CH4 and Group II: HF
Answer:
Group I: SnH4 and Group II: HF
Explanation:
In the case of group 14 hydrides, we know that the melting and boiling points of compounds increase down the group. Since the melting and boiling points of compounds increase down the group, SnH4 will have the highest boiling point.
In the case of group 17 hydrides, we know that electro negativity decreases down the group. As electro negativity decreases, so does the magnitude of hydrogen bonding among group 17 hydrides. Hence, HF has the highest boiling point among group 17 hydrides since it has the greatest magnitude of hydrogen bonding between its molecules.
how many cg are equal to 0.459kg
1kg = 100000
0.459kg = 0.459*100000
= 45900cg
→ 0.459kg = 45900cg
According to unit conversion and as 1 kg=100,000 cg , there are 45900 cg in 0.459 kg.
What is unit conversion?Unit conversion is defined as a multi-step process which involves multiplication or a division operation by a numerical factor.The process of unit conversion requires selection of appropriate number of significant figures and the rounding off procedure.
It involves a conversion factor which is an expression for expressing the relationship between the two units.A conversion ratio always has value which equals to one which indicates that numerator and denominator have values which are expressed in different units.
It is an easy process which involves unit conversion between the different conversion systems. While unit conversion care needs to be taken regarding accuracy and precision.
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The next two questions provide some more practice on calculations using half-lives. The isotope 64Cu has t1/2 of 12.7 hours. If the initial concentration of this isotope in an aqueous solution is 845 ppm, what will the abundance be after 4.00 hours? To solve this problem, first use equation (7) to determine k for 64Cu; then use this k value in equation (6) to obtain the amount of 64Cu, A, remaining after 4.00 hours if the amount present at the start, A0, is 845 ppm.
Answer:
The value is [tex] A = 679.5 \ ppm[/tex]
Explanation:
From the question we are told that
The half life of [tex]^{64}Cu[/tex] is [tex] t_h = 12.7 \ hr [/tex]
The initial concentration is [tex] A_o = 845 \ ppm [/tex]
The time duration is [tex] t = 4 \ hr [/tex]
Generally the rate constant is mathematically represented as
[tex] k = \frac{0.693}{t_h} [/tex]
[tex] k = \frac{0.693}{12.7} [/tex]
[tex] k = 0.0545 \ hr^{-1} [/tex]
This rate constant is also mathematically represented as
[tex] k = \frac{1}{t} * ln (\frac{A_o}{A}) [/tex]
Here A is the remaining concentration after t
So
[tex] 0.0545 = \frac{1}{4} * ln (\frac{845}{A}) [/tex]
[tex] 0.218 = ln (\frac{845}{A}) [/tex]
[tex] e^{0.218} = \frac{845}{A} [/tex]
[tex] 1.2436 = \frac{845}{A} [/tex]
[tex] A = \frac{845}{1.2436} [/tex]
[tex] A = 679.5 \ ppm[/tex]
