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2 years ago

12

Given the following equation: 4 NH3 (g)5 O2 (g) > 4 NO (g) 6 H20 () + How many moles of H20 is produced if 0.44 mol of NH3 re

acts?

1 answer:

ExtremeBDS [4]2 years ago

7 0

Answer : The number of moles of $H_2O$ produced are, 0.66 mole.

Explanation : Given,

Given moles of $NH_3$ = 0.44 mole

The given balanced chemical reaction is,

$4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)$

From the given balanced chemical reaction, we conclude that

As, 4 moles of $NH_3$ react to give 6 moles of $H_2O$

So, 0.44 moles of $NH_3$ react to give $\frac{6}{4}\times 0.44=0.66$ moles of $H_2O$

Therefore, the number of moles of $H_2O$ produced are, 0.66 mole.

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What is the number of moles in 15.0 g AsH3?

pentagon [3]

We are given with the mass of Arsine ( $AsH_{3}$

The mass of arsine is 15g

there is a relation between moles, mass and molar mass of any compound which is

$moles=\frac{mass}{molarmass}$

The molar mass of Arsine = atomic mass of As + 3X atomic mass of H

the molar mass of Arsine = 74.92 + 3X 1 = 77.92 g/mol

Let us calculate the moles as

$moles=\frac{15}{77.92}=0.192mol$

7 0

2 years ago

A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg

Brut [27]

Answer: $1.0\times 10^2g$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(-6.4-(13.6))^0C=7.2^0C$ = Depression in freezing point

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

$K_f$ = freezing point constant = $3.96^0C/m$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}}\times \text{weight of solvent in kg}}$

Weight of solvent (X)= 950 g = 0.95 kg

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

$7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}$

$x=1.0\times 10^2g$

Thus $1.0\times 10^2g$ urea was dissolved.

8 0

3 years ago

86.14 mL of an acid solution was needed to neutralize 30.24 mL of a base solution of unknown concentrations. A second trial is r

Airida [17]

Answer:

The correct answer is option B.

Explanation:

As given ,that 30.24 mL of base was neutralize by 86.14 mL of acid which means that moles of base present in 30.24 mL are neutralized by moles of acid present in 86.14 mL.

After dilution of base from 30.24 mL to 50.0 mL .Since, the moles of base are same in the solution as that of the moles in solution before dilution. Moles of acid require to neutralize the base after dilution will same as a that of present moles of acid present in 86.14 mL.

5 0

2 years ago

Why is the anatomy of the human body not a general reason to study chemistry. Need help for a good reply.

ivolga24 [154]

Anatomy's a biology topic!

8 0

2 years ago

When a hydrogen atom makes the transition from the second excited state to the ground state (at -13.6 eV) the energy of the phot

viktelen [127]

Answer : The energy of the photon emitted is, -12.1 eV

Explanation :

First we have to calculate the $'n^{th}'$ orbit of hydrogen atom.

Formula used :

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number of hydrogen atom = 1

Energy of n = 1 in an hydrogen atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6eV$

Energy of n = 2 in an hydrogen atom:

$E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV$

Energy change transition from n = 1 to n = 3 occurs.

Let energy change be E.

$E=E_-E_3=(-13.6eV)-(-1.51eV)=-12.1eV$

The negative sign indicates that energy of the photon emitted.

Thus, the energy of the photon emitted is, -12.1 eV

3 0

2 years ago