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tekilochka [14]
3 years ago
5

I WILL GIVE BRAINLIEST!!!!!

Chemistry
2 answers:
Akimi4 [234]3 years ago
7 0

Answer:

D is your answer hope this helps

Explanation:

LenKa [72]3 years ago
4 0
D. Electron cloud allowed the particles to pass through
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_______ is force divided by area.
Digiron [165]

Answer:

Pressure

Explanation:

Basically when you are pressing something you are applying force over an area thus Pressure is force divided by area.

7 0
3 years ago
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True or false atoms are made up of tiny particles called molecules true or false
Nana76 [90]
False, Molecules are made of atoms and atoms are made of quarks.
4 0
3 years ago
A student reduces the temperature of a 300 cm3 balloon from 60°c to 30°
FromTheMoon [43]

We can use the ideal gas equation which is expressed as PV = nRT. At a constant pressure and number of moles of the gas the ratio T/V is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

T1 / V1 = T2 / V2

V2 = T2 x V1 / T1

V2 = 303.15  x 300 / 333.15

<span>V2 = 272.99 cm³</span>

7 0
4 years ago
Combustion reactions are exothermic. The heat of reaction for the combustion of 2-methylheptane, C8H18, is 1.306×103 kcal/mol. W
WARRIOR [948]

Answer:

11.45kcal/g

2.612 × 10³ kcal

Explanation:

When a compound burns (combustion) it produces carbon dioxide and water. The combustion of 2-methylheptane can be represented by the following balanced equation:

2 C₈H₁₈ + 25 O₂ ⇄ 16 CO₂ + 18 H₂O

It releases  1.306 × 10³ kcal every 1 mol of C₈H₁₈ that is burned.

<em>What is the heat of combustion for 2-methylheptane in kcal/gram?</em>

We know that the molar mass of C₈H₁₈ is 114.0g/mol. Then, using proportions:

\frac{1.306 \times 10^{3}Kcal}{1mol} .\frac{1mol}{114.0g} =11.45kcal/g

<em>How much heat will be given off if molar quantities of 2-methylheptane react according to the following equation? 2 C₈H₁₈ + 25 O₂ ⇄ 16 CO₂ + 18 H₂O</em>

In this equation we have 2 moles of C₈H₁₈. So,

2mol \times\frac{1.306 \times 10^{3}kcal }{1mol} =2.612\times 10^{3}kcal

3 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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