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Tanzania [10]
3 years ago
10

Find the molar mass of nitric acid (HNO3).

Chemistry
1 answer:
Pachacha [2.7K]3 years ago
7 0

Answer:

M = 63.02g/mol

Explanation:

Molar mass = the atomic masses of each atom combined

H = 1.01

N = 14.01

O = 16.00

M = 1.01(1) + 14.01(1) + 16.00(3)

M = 63.02

Therefore the molar mass of nitric acid is 63.02g/mol

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How many milliliters of 0.25M H2SO4 can be prepared from 57 mL of a 3.0M solution of H2SO4?
DochEvi [55]

Answer:

Why ? Because 1 molecule of H2SO4 gives 2 H+ ions per molecule while only one H+ ion is required to neutralize 1 molecule of KOH. So, 1 molecule of H2SO4 can neutralize 2 molecules of KOH. Hence, we would require 525 ml of 0.03 M H2SO4 to neutralize 525 ml of 0.06 M KOH. How will we prepare 525 ml of 0.03 M H2SO4 ?

Explanation:

Now, we have 0.025 M H2SO4 and we do not know how much volume we have.

We will use the standard N1 X V1 = N2 X V2 for this calculation.

N1=0.025 M; V1=unknown; N2=0.03 M and V2=525 ml.

So V1= (0.03 X 525)/(0.025) = 630 ml.

5 0
3 years ago
Why do atoms always have an overall neutral charge?
PSYCHO15rus [73]
This means the nucleus of an atom is always positively charged
5 0
3 years ago
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What is the frequency of green light that has a wavelength of 5.14 x 10^-7m?
nata0808 [166]

Answer:

Frecuency = 5,83x10⁻⁷ Hz

Explanation:

The equation that connects wavelenght and frequency is given by:

λ = c/ν

λ=wavelenght (expressed in lenght´s units)

c= speed of light (3x10⁸ m/sec)

ν=frequency (expressed in units of time⁻¹ or Herzt)

In our case, λ=5,14x10⁻⁷ m , so replacing in our previous formula, this gives us the final result of ν (frequency for green light) of 5,83x10¹⁴ Hz (or Herzt)

5 0
3 years ago
The table below gives the equilibrium concentrations for this reaction at a certain temperature : N2(g) + O2(g) 2NO(g) what is t
makkiz [27]
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A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure
Lesechka [4]

Answer:

the final volume of the gas is V_2 = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure P_{ext}:

P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}

P_{ext} = 1.03 \ atm

The workdone W = P_{ext}V

The change in volume ΔV= \dfrac{W}{P_{ext}}

ΔV = \dfrac{130.1 \ J  \times \dfrac{1 \ L  \ atm}{ 101.325 \ J}  }{1.03 \ atm }

ΔV = \dfrac{1.28398717 }{1.03  }

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial  volume = 61.5 mL

The change in volume V is \Delta V = V_2 -V_1

-  V_2= -  \Delta V  -V_1

multiply through by (-), we have:

V_2=   \Delta V+V_1

V_2 =  1250 mL + 61.5 mL

V_2 = 1311.5 mL

∴ the final volume of the gas is V_2 = 1311.5 mL

5 0
3 years ago
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