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3 years ago

16) The mass number of an element whose atoms have 12 protons and 13 neutrons is_

Chemistry

1 answer:

Bess [88]3 years ago

6 0

The answer is c) 25 because the mass number is the # of protons and neutrons in the nucleus, therefor, 12 + 13 = 25

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2(x-1 ➗ x+3)-7(x+3 ➗ x-1)=5

NeTakaya

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3 years ago

compare the boiling point of water at sea leavel to the boiling point in denver (altitude = 1 mile) Explain

steposvetlana [31]

Answer:

boiling point decreases in denver

Explanation:

in higher places

theres less atmospheric pressure

it takes less energy to bring water to the boiling point.

Less energy means less heat

which means water will boil at a lower temperature

wonderopolisorg

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1 year ago

What change in energy occurs when you burn coal or petroleum??

expeople1 [14]

Chemical energy is converted to thermal energy.

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2 years ago

Read 2 more answers

A sample of a pure compound that weighs 59.8 g contains 27.6 g Sb (antimony) and 32.2 g F (fluorine). What is the percent compos

Rama09 [41]

Answer:

53.85%

Explanation:

Data obtained from the question include:

Mass of antimony (Sb) = 27.6g

Mass of Fluorine (F) = 32.2g

Mass of compound = 59.8g

Percentage composition of fluorine (F) =..?

The percentage composition of fluorine can be obtained as follow:

Percentage composition of fluorine = mass of fluorine/mass of compound x 100

Percentage composition of fluorine = 32.2/59.8 x 100

= 53.85%

Therefore, the percentage composition of fluorine in the compound is 53.85%

3 0

3 years ago

For the reaction 2 H2S(g) D 2 H2 (g) + S2 (g), Kp = 1.5 × 10−5 at 800.0°C. If the initial partial pressures of H2 and S2 in a cl

Usimov [2.4K]

Answer: The approximate equilibrium partial pressure of $H_2S$ is 3.92 atm

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

$2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)$

$K_p=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}$

$1.5\times 10^{-5}=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}$

On reversing the reaction:

$2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)$

initial pressure 4.00atm 2.00 atm 0

eqm (4.00-2x)atm (2.00-x) atm 2x atm

$K_p=\frac{[H_2S]^2}{[H_2]^2\times [S_2]}$

$K_p'=\frac{1}{K_p}=0.67\times 10^5$

$2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)$

$0.67\times 10^5=\frac{2x]^2}{[4.00-2x]^2\times [2.00-x]}$

$x=1.96$

$[H_2S]=2x=2\times 1.96=3.92 atm$

Thus approximate equilibrium partial pressure of $H_2S$ is 3.92 atm

3 0

3 years ago