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max2010maxim [7]
2 years ago
13

Which of the following factors does not affect a reaction rate?

Chemistry
1 answer:
Natalka [10]2 years ago
4 0
The correct answer is C
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Which of the following statements is TRUE about the loose rocks on the surface
Ivan

Answer:B

Explanation:

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3 years ago
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The empirical formula of glucose is CH2O, and the molecular formula of glucose is 6 times than the empirical formula. What is th
goldfiish [28.3K]
C6H12O6
That is the molecular formula of glucose
3 0
3 years ago
A student obtains a mixture of the chlorides of two unknown metals, X and Z. The percent by mass of X and the percent by mass of
Aleksandr-060686 [28]

<u>Answer:</u> The additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X

<u>Explanation:</u>

To calculate the mole percent of a substance, we use the equation:

\text{Mole percent of a substance}=\frac{\text{Moles of a substance}}{\text{Total moles}}\times 100

Mass percent means that the mass of a substance is present in 100 grams of mixture

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We require the molar masses of Z and X to calculate the mole percent of Z and X respectively

Hence, the additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X

8 0
3 years ago
If air is 80 percent nitrogen and 20 percent oxygen by volume what is density
Evgen [1.6K]
1 mol of any gas or mix of gases at STP conditions will have a volume of 22.4 L. Since the problem doesn’t said what are the conditions I will asume that are STP condition and the volume of one mole of the mix will have a volume of 22.4 L.

You may know that density is
D=m/v

In one mole of air I will have 80% of Nitrogen (N2) and 20% oxygen (O2).

So the mass of one mole of air will be

14 x2x0.80+16x2x0.20 = 22.4 g + 6.4 g = 28.8 g

D= 28.8/22.4 = 1.28 g/L

Of course if the temperature is higher the density will be smaller because the volume of one mole will be bigger and viceversa if the temperature decrease. Also if the pressure is different than one atm the volume of a mol will change.
5 0
3 years ago
RATE LAW QUESTION !
vivado [14]
In general, we have this rate law express.:

\mathrm{Rate} = k \cdot [A]^x [B]^y
we need to find x and y

ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).

then we go to compare two experiments in which only one concentration is changed

compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4)  by the smaller [B] (experiment 1) and call it Δ[B]

Δ[B]= 0.3 / 0.1 = 3

now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:

ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...

solve for y in the equation \Delta \mathrm{Rate} = \Delta [B]^y

3.09 = (3)^y \implies y \approx 1

To this point, \mathrm{Rate} = k \cdot [A]^x [B]^1

do the same to find x.
choose two experiments in which only the concentration of B is unchanged:

Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4

solve for x for \Delta \mathrm{Rate} = \Delta [A]^x

4=  (2)^x \implies x = 2

the rate law is

Rate = k·[A]²[B]
6 0
2 years ago
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