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Umnica [9.8K]
3 years ago
8

Values for the molar mass of nitrogen, oxygen, and nitrogen dioxide molecules are given in the table below. What mass of nitroge

n dioxide molecules is formed when 1 mole of nitrogen reacts with 2 moles of oxygen and forms nitrogen dioxide? Choices: A). 92.02g B). 46.01g C). 23.00g D) 2.00g
Chemistry
1 answer:
Gnoma [55]3 years ago
4 0

Answer:

A). 92.02g

Explanation:

Equation of the reaction;

N2 (g)+ 2O2(g)------> 2NO2(g)

Note that the balanced reaction equation is the first step in solving any problem on stoichiometry. Once the reaction equation is correct, the question can be easily solved.

Reaction of one mole of nitrogen gas with two moles of oxygen gas yields two moles of nitrogen dioxide.

Mass of two moles of nitrogen dioxide= 2[14 + 2(16)] = 2[14+32]= 2[46]= 92 gmol-1

Therefore; Mass of two moles of nitrogen dioxide is 92

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The volume of copper : 3.24 ml

<h3>Further explanation </h3>

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than objects with a smaller type of mass

The unit of density can be expressed in g/cm³ or kg/m³

Density formula:

\large{\boxed {\bold {\rho~=~ \frac {m} {V}}}}

ρ = density , g/cm³ or kg/m³

m = mass , g or kg

v = volume , cm³  or m³

A common example is the water density of 1 gr / cm³

The density of copper : 8.96 gr/ml

mass of copper : 29 g

then the volume :

\tt V=\dfrac{m}{\rho}\\\\V=\dfrac{29~g}{8.96~g/ml}\\\\V=\boxed{\bold{3.24~ml~or~3.24~cm^3}}

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3 years ago
How have ideas about the atom changed in the last two centureies?
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A square block of steel with volume 10 cm3 and mass of 75 g is cut precisely in half. The density of the two smaller pieces is n
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It will be the exact same density
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GalinKa [24]

Answer:

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3 0
3 years ago
When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°
dlinn [17]

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.30^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

K_f = freezing point constant =  

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8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}

x=58.2g

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

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