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Oduvanchick [21]
3 years ago
7

The following thermochemical equation is for the reaction of nitrogen(g) with oxygen(g) to form nitrogen dioxide(g). N2(g) + 2O2

(g)2NO2(g) H = 66.4 kJ How many grams of N2(g) would be made to react if 14.2 kJ of energy were provided?
Chemistry
1 answer:
Rzqust [24]3 years ago
8 0

<u>Answer:</u> The mass of nitrogen gas reacted to produce given amount of energy is 5.99 grams.

<u>Explanation:</u>

The given chemical reaction follows:

N_2(g)+2O_2(g)\rightarrow 2NO_2(g);\Delta H=66.4kJ

We know that:

Molar mass of nitrogen gas = 28 g/mol

We are given:

Enthalpy change of the reaction = 14.2 kJ

To calculate the mass of nitrogen gas reacted, we use unitary method:

When enthalpy change of the reaction is 66.4 kJ, the mass of nitrogen gas reacted is 28 grams.

So, when enthalpy change of the reaction is 14.2 kJ, the mass of nitrogen gas reacted will be = \frac{28}{66.4}\times 14.2=5.99g

Hence, the mass of nitrogen gas reacted to produce given amount of energy is 5.99 grams.

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crimeas [40]

Answer:

The answer is below

Explanation:

A normal model is represented as (μ, σ). Therefore for (1.5, 0.18), the mean (μ) = 1.5 and the standard deviation (σ) = 0.18

The z score shows by how many standard deviations the raw score is above or below the mean. It is given as:

z=\frac{x-\mu}{\sigma}

a) For x < 1.35 s

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.35-1.5}{0.18}=-0.83

From the normal distribution table, the percent of drivers have a reaction time less than 1.35 seconds = P(x < 1.35) = P(z < -0.83) = 0.2033 = 20.33%

b) For x > 1.9 s

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.9-1.5}{0.18}=2.22

From the normal distribution table, the percent of drivers have a reaction time greater than 1.9 seconds = P(x > 1.9) = P(z > 2.22) = 1 - P(z<2.22) = 1 - 0.9868 = 0.0132 = 1.32%

c) For x = 1.45

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.45-1.5}{0.18}=-0.28

For x = 1.75

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.75-1.5}{0.18}=1.39

From the normal distribution table, P(1.45 < x < 1.75) = P(-0.28 < z < 1.39) = P(z < 1.39) - P(z< - 0.28) = 0.9177 - 0.3897 = 0.528 = 52.8%

d) A percentage of 10% corresponds to a z score of -1.28

z=\frac{x-\mu}{\sigma}\\\\-1.28=\frac{x-1.5}{0.18}\\\\x-1.5=-0.2034\\\\x=1.27

e) P(z < z1) - P(z< -z1) = 60%

P(z < z1) - P(z< -z1) = 0.6

P(z < -z1) = 1 - P(z < z1)

P(z<z1) - (1 - P(z < z1)) = 0.6

2P(z<z1) - 1= 0.6

2P(z<z1) = 1.6

P(z<z1) = 0.8

From the z table, z1 = 0.85

0.85=\frac{x-1.5}{0.18}and-0.85=\frac{x -1.5}{0.18}  \\\\x=1.65 \ and\ x=1.35

The reaction time between 1.35 and 1.65 seconds

8 0
3 years ago
What can u say about the sizes of all the atoms in a piece of carbon​
vlada-n [284]

Answer:

all atoms must have 6 protons to be a carbon atom

Explanation:

brainliest?

4 0
2 years ago
Determine the pH of a 0.048 M hypochlorous acid (HClO) solution. Hypochlorous acid is a weak acid (Ka = 4.0 ✕ 10−8 M).
storchak [24]

pH of 0.048 M HClO is 4.35.

<u>Explanation:</u>

HClO is a weak acid and it is dissociated as,

HClO ⇄ H⁺ + ClO⁻

We can write the equilibrium expression as,

Ka = $\frac{[H^{+}] [ClO^{-}]  }{[HClO]}

Ka = 4.0 × 10⁻⁸ M

4.0 × 10⁻⁸ M = $\frac{x \times x }{0.048}

Now we can find x by rewriting the equation as,

x² =  4.0 × 10⁻⁸ × 0.048

   = 1.92 × 10⁻⁹

Taking sqrt on both sides, we will get,

x = [H⁺] = 4.38 × 10⁻⁵

pH = -log₁₀[H⁺]

     = - log₁₀[ 4.38 × 10⁻⁵]

   = 4.35

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2 years ago
Osmium is the most dense element we know of. A 22 g sample of Osmium has a volume of 100 cL. Calculate the density of Osmium in
kakasveta [241]

Answer:

a. V = 1000 mL

b. Denisty = 0.022 g/mL

Explanation:

a.

First we need to convert the volume of the Osmium into mL. For that purpose we are given the conversion unit as:

1 mL = 0.1 cL

Hence, the given volume of Osmium will be:

V = Volume of Osmium = 100 cL = (100 cL)(1 mL/0.1 cL) = 1000 mL

<u>V = 1000 mL</u>

b.

The density of Osmium is given by the following formula:

Density = mass/Volume

Denisty = 22 g/1000 mL

<u>Denisty = 0.022 g/mL</u>

5 0
3 years ago
Mg + 2HCl → MgCl2 + H2
deff fn [24]

Answer:

Explanation:

This is a limiting reactant problem.

Mg(s)

+

2HCl(aq)

→

MgCl

2

(

aq

)

+ H

2

(

g

)

Determine Moles of Magnesium

Divide the given mass of magnesium by its molar mass (atomic weight on periodic table in g/mol).

4.86

g Mg

×

1

mol Mg

24.3050

g Mg

=

0.200 mol Mg

Determine Moles of 2M Hydrochloric Acid

Convert

100 cm

3

to

100 mL

and then to

0.1 L

.

1 dm

3

=

1 L

Convert

2.00 mol/dm

3

to

2.00 mol/L

Multiply

0.1

L

times

2.00 mol/L

.

100

cm

3

×

1

mL

1

cm

3

×

1

L

1000

mL

=

0.1 L HCl

2.00 mol/dm

3

=

2.00 mol/L

0.1

L

×

2.00

mol

1

L

=

0.200 mol HCl

Multiply the moles of each reactant times the appropriate mole ratio from the balanced equation. Then multiply times the molar mass of hydrogen gas,

2.01588 g/mol

0.200

mol Mg

×

1

mol H

2

1

mol Mg

×

2.01588

g H

2

1

mol H

2

=

0.403 g H

2

0.200

mol HCl

×

1

mol H

2

2

mol HCl

×

2.01588

g H

2

1

mol H

2

=

0.202 g H

2

The limiting reactant is

HCl

, which will produce

0.202 g H

2

under the stated conditions.

pls mark as brainliest ans

7 0
2 years ago
Read 2 more answers
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