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3 years ago

How many molecules are there in 7.27 miles of ammonia?

Chemistry

2 answers:

worty [1.4K]3 years ago

5 0

Answer:

43.78 X 10^23

Explanation:

1 mole of NH3 has 6.023×10^23 molecules according to avogadro's principle

therefore, 7.27 moles = 6.023×10^23 X 7.27

= 43.78 X 10^23

NemiM [27]3 years ago

4 0

Hey There!

To solve this question remember that 1 mole of ammonia is equal to 17G of ammonia.

Using that, we first multiply 17 by 7 = 24

Then we multiply 7 by .27 = 1.89

Finally, taking those and adding them together gives you the answer...

25.89 grams :)

Have a great day, I hope I helped.

+ Brainliest & Five stars are always appreciated

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Meat-eating animals are:

Rina8888 [55]

Answer:

second-order consumers.

Explanation:

carnivores and omnivores are secondary consumers. many carnivores eat herbivores. some eat omnivores and some eat other carnivores. carnivores that consume other carnivores are called tertiary consumers.

8 0

2 years ago

Read 2 more answers

Use the following information S(s)+O2(g)→SO2(g), ΔH∘ = -296.8kJ SO2(g)+12O2(g)→SO3(g) , ΔH∘ = -98.9kJ to calculate ΔH∘f for H2SO

Irina-Kira [14]

Answer : The enthalpy of formation of $H_2SO_4$ is, -812.4 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation of $H_2SO_4$ will be,

$S+H_2+2O_2\rightarrow H_2SO_4$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $S+O_2\rightarrow SO_2$ $\Delta H_1=-298.2$

(2) $SO_2+\frac{1}{2}O_2\rightarrow SO_3$ $\Delta H_2=-98.2$

(3) $SO_3+H_2O\rightarrow H_2SO_4$ $\Delta H_3=-130.2$

(4) $H_2+\frac{1}{2}O_2\rightarrow H_2O$ $\Delta H_4=-285.8$

Now adding all the equations, we get the expression for enthalpy of formation of $H_2SO_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4$

$\Delta H=(-298.2)+(-98.2)+(-130.2)+(-285.8)$

$\Delta H=-812.4kJ/mol$

Therefore, the enthalpy of formation of $H_2SO_4$ is, -812.4 kJ/mole

3 0

3 years ago

How many moles are in 5.0 x 10 *23 atoms of iron?

tamaranim1 [39]

1 mole ------------ 6.02 x 10²³ atoms
? moles ----------- 5.0 x 10²³ atoms

moles = 5.0 x 10²³ / 6.02 x 10²³

= 0.830 moles

hope this helps!

7 0

3 years ago

What is the oxidation number of chromium in dichromate ion, cr2o72-?

castortr0y [4]

Oxidation number is charge of element in compound. Can be neutral, positive or negative.
Oxygen in dichromate has oxidation number -2, becauce there are seven oxygens, net oxidation number of oxygen is -14.
2·x(oxidation number of Cr) + 7· (-2) = -2 2x= +12
x= +6, oxidation number of one chromium is +6.

8 0

3 years ago

When the temperature of a rigid hollow sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 1.89 x

Leno4ka [110]

You have to use the equation PV=nRT.
P=pressure (in this case 1.89x10^3 kPa which equals 18.35677 atm)
1V=volume (in this case 685L)
n=moles (in this case the unknown)
R=gas constant (0.08206 (L atm)/(mol K))
T=temperature (in this case 621 K)
with the given information you can rewrite the ideal gas law equation as n=PV/RT.
n=(18.35677atm x 685L)/(0.08206atmL/molK x 621K)
n=246.8 moles

8 0

2 years ago