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3 years ago

which of the following is a product formed whrb NO2 and H2O react together A O2 B HNO3 C H2 D P(OH)2

Chemistry

2 answers:

marshall27 [118]3 years ago

8 0

The answer to your question is HNO3

tino4ka555 [31]3 years ago

8 0

Answer : The correct option is, (B) $HNO_3$

Explanation :

Balanced chemical reaction : It is defined as the number of atoms of individual elements present on reactant side must be equal to the product side.

The balanced chemical reaction will be,

$3NO_2+H_2O\rightarrow 2HNO_3+NO$

By the stoichiometry, we can say that 2 moles of $NO_2$ react with 1 mole of $H_2O$ to give 1 moles of $HNO_3$ and 1 mole of $NO$ .

In this reaction, $NO_2$ and $H_2O$ are the reactants and $HNO_3$ and $NO$ are the products.

Hence, the correct option is, (B) $HNO_3$

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Which of the following transition is considered the phase<br> change from water to steam?

Alexandra [31]

Answer:

the transition from water to steam is call evaporation.

Explanation:

water evaporates or vaporizes and turns into steam this typically happens when heat meets with the water. an example would be if you boil water steam will release from the pot, or when it rains, snows, etc (precipitation) the water evaporates and the ground dries.

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3 years ago

#1: Which of the following scientists came up with the first widely recognized atomic theory?

boyakko [2]

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4 years ago

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Law Incorporation [45]

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3 years ago

Enter your answer in the box provided. How many grams of helium must be added to a balloon containing 6.24 g helium gas to doubl

Archy [21]

Answer : The mass of helium gas added must be 12.48 grams.

Explanation : Given,

Mass of helium (He) gas = 6.24 g

Molar mass of helium = 4 g/mole

First we have to calculate the moles of helium gas.

$\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}=\frac{6.24g}{4g/mole}=1.56moles$

Now we have to calculate the moles of helium gas at doubled volume.

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

$V\propto n$

or,

$\frac{V_1}{V_2}=\frac{n_1}{n_2}$

where,

$V_1$ = initial volume of gas = V

$V_2$ = final volume of gas = 2V

$n_1$ = initial moles of gas = 1.56 mole

$n_2$ = final moles of gas = ?

Now we put all the given values in this formula, we get

$\frac{V}{2V}=\frac{1.56mole}{n_2}$

$n_2=3.12mole$

Now we have to calculate the mass of helium gas at doubled volume.

$\text{Mass of }He=\text{Moles of }He\times \text{Molar mass of }He$

$\text{Mass of }He=3.12mole\times 4g/mole=12.48g$

Therefore, the mass of helium gas added must be 12.48 grams.

4 0

3 years ago

Potassium hydroxide is used to precipitate each of the cations from their respective solution. determine the minimum concentrati

MAXImum [283]

This is the three cases that help to determine the minimum concentration of KOH required for precipitation
Part a) 1.5×10^−2 M K CaCl2
Part b) 2.3×10^−3 M Fe (NO3)2
Part c) 2.0×10^−3 M MgBr2

a) CaCl2 + 2KOH --> Ca (OH) 2 + 2KCl Ca (OH) 2 <=> Ca^2+ + 2OH^-
ksp = 1.5*10^-2 + x^2
4.68*10^-6 = 1.5*10^-2 + x^2
x= [KOH] = 0.01766

b) Fe (NO3)2 +2 KOH--> Fe (OH)2 + 2KNO3
Fe (OH)2 <=> Fe^2+ + 2OH^-
ksp = 2.3*10^-3 + x^2
4.87*10^-17 = 2.3*10^-3 + x^2
x= 1.46*10^-7

c) MgBr2 + KOH --> Mg (OH) 2 + 2KBr
Mg (OH) 2 <=> Mg^2+ + 2OH^-
ksp = 2.0*10^-3 + x^2
2.06*10^-13 = 2.0*10^-3 + x^2
x= 1.015*10^-5

8 0

4 years ago