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Usimov [2.4K]
2 years ago
11

Identify the total number of molecules in the chemical formula: 6CH4 a 6 b 10 c 24 d 30

Chemistry
1 answer:
Anon25 [30]2 years ago
8 0

Answer: A.6

Explanation:

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3 years ago
Which of the following is true?
tresset_1 [31]

Answer:

B

Explanation:

The sun can be dangerous but it also helps our Earth in a lot of ways

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How much does the mass of 12.65-g sample of copper(II) nitrate hexahydrate decrease when heated?
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There is 103 percent of water in hydrate
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2 years ago
Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. At what radius (in nm) of sphe
Gala2k [10]

Answer:

Radius of the interior sphere = 3.847 nm

Explanation:

The volume of the shell (Vs) is equal to the difference of the volume of the outer sphere (Vo) and the volume of the inner sphere (Vi). Then:

V_s=V_o-V_i=V_i\\V_o=2*V_i\\

If we express the radius of the outer sphere (ro) in function of the radius of the inner sphere (ri), we have (e being the shell thickness):

r_o=r_i+e

The first equation becomes

\frac{4 \pi}{3}*r_o^{3}  = 2 * \frac{4 \pi}{3}*r_i^{3}  \\\\\frac{4 \pi}{3}*(r_i+e)^{3}  = 2 * \frac{4 \pi}{3}*r_i^{3}  \\\\(r_i+e)^{3}  = 2 *r_i^{3}  \\\\(r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3})=2r_i^{3}\\\\-r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3}=0\\\\-r_i^{3}+3r_i^{2}+3r_i+1=0

To find ri that satisfies this equation we have to find the roots of the polynomial.

Numerically, it could be calculated that ri=3.847 nm satisfies the equation.

So if the radius of the interior sphere is 3.847 nm, the volume of the interior sphere is equal to the volume of the shell of 1nm.

8 0
3 years ago
The average score for games played in the NFL is 21.1 and the standard deviation is 8.9 points. 46 games are randomly selected.
kaheart [24]

Answer:

A) P(21.4317 < ¯x < 22.7561) = 0.2975

B) Q1 for the ¯x distribution = 21.9844

Explanation:

The Central Limit theorem allows us to say that

Sample mean = Population mean = 21.1 points

Mean of sampling distribution = σₓ = (σ/√n)

σ = population standard deviation = 8.9 points

n = sample size = 46

σₓ = (8.9/√46) = 1.3122334098 = 1.3122

A) P(21.4317 < ¯x < 22.7561) =

This is a normal distribution problem

To find this probability, we will use the normal probability tables

We first normalize/standardize 21.4317 and 22.7561.

The standardized score of any value is that value minus the mean divided by the standard deviation.

For 21.4317

z = (x - μ)/σ = (21.4317 - 21.1)/1.3122 = 0.25

For 22.7561

z = (x - μ)/σ = (22.7561 - 21.1)/1.3122 = 1.26

The required probability

P(21.4317 < ¯x < 22.7561) = P(0.25 < z < 1.26)

Checking the tables

P(21.4317 < ¯x < 22.7561) = P(0.25 < z < 1.26)

= P(z < 1.26) - P(z < 0.25)

= 0.89617 - 0.59871

= 0.29746 = 0.2975 to 4 d.p.

B) Q1 for the distribution is the first quartile. The first quartile is greater than 25% of the distribution.

P(x > Q1) = 0.25

Let the z-score that corresponds to Q1 be z'

P(x > Q1) = P(z > z') = 0.25

But P(z > z') = 1 - P(z ≤ z') = 0.25

P(z ≤ z') = 1 - 0.25 = 0.75

From the normal distribution tables,

z' = 0.674

z' = (Q1 - μ)/σ

0.674 = (Q1 - 21.1)/1.3122

Q1 = 0.674×1.3122 + 21.1 = 21.9844228 = 21.9844 to 4 d.p.

Hope this Helps!!!

7 0
3 years ago
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