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Naddik [55]
3 years ago
9

2) A 0.77 mg sample of nitrogen gas reacts with chlorine gas to form 6.61 mg of a nitrogen

Chemistry
1 answer:
tankabanditka [31]3 years ago
8 0

Answer:

NCl₃

Explanation:

From the question given above, the following data were obtained:

Mass of nitrogen (N) = 0.77 mg

Mass of chlorine (Cl) = 6.61 mg

Empirical formula =?

The empirical formula of the compound can be obtained as follow:

N = 0.77 mg

Cl = 6.61 mg

Divide by their molar mass

N = 0.77 / 14 = 0.055

Cl = 6.61 / 35.5 = 0.186

Divide by the smallest

N = 0.055 / 0.055 = 1

Cl = 0.186 /0.055 = 3

Therefore, the empirical formula of the compound is NCl₃

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How many particles are there in 5.0 moles​
grigory [225]

Answer:

3.01 × 10^24 particles

Explanation:

According to Avagadro, in one mole of a substance, there are 6.02 × 10^23 atoms or particles.

Using the formula: N = n × NA

Where;

N= number of particles or atoms

n = number of moles

NA = Avagadro's constant or number

This means that for 5 moles of a substance, there will be:

5 × 6.02 × 10^23

= 30.1 × 10^23

= 3.01 × 10^24 particles

8 0
3 years ago
What is the vapor pressure of CS2CS2 in mmHgmmHg at 26.5 ∘C∘C? Carbon disulfide, CS2CS2, has PvapPvap = 100 mmHgmmHg at −−5.1 ∘C
Digiron [165]

Answer: 26.5 mm Hg

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1= initial pressure at 26.5^oC = ?

P_2 = final pressure at -5.1^oC = 100 mm Hg

= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol

R = gas constant = 8.314 J/mole.K

T_1= initial temperature = 26.5^oC=273+26.5=299.5K

T_2 = final temperature =-5.1^oC=273+(-5.1)=267.9K

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

\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}]

\log  (\frac{P_1}{100})=-0.576

\frac{P_1}{100}=0.265

P_1=26.5mmHg

Thus the vapor pressure of CS_2CS_2 in mmHg at 26.5 ∘C is 26.5

7 0
3 years ago
A 5-kg quantity of radioactive isotope decays to 2 kg after 10 years. Find the decay constant of the isotope.k = _____
Dvinal [7]

Answer:

k = -0.09165 years^(-1)

Explanation:

The exponential decay model of a radioactive isotope is generally given as;

A(t) = A_o(e^(kt))

Where;

A_o is quantity of isotope before decay, k is decay constant and A(t) is quantity after t years

We are given;

A_o = 5 kg

A(10) = 2kg

t = 10 years

Thus;

A(10) = 2 = 5(e^(10k))

Thus;

2 = 5(e^(10k))

2/5 = (e^(10k))

0.4 = (e^(10k))

In 0.4 = 10k

-0.9164 = 10k

k = -0.9164/10

k = -0.09165 years^(-1)

7 0
2 years ago
A neutral atom of gold has the same number of electrons and what?
liraira [26]
Atomic mass!!! the electrons and atomic mass will always be the same most the time
6 0
3 years ago
Determine which choice is an example of an endothermic process.
klemol [59]

Answer:

D. Baking bread

Explanation:

In this process, energy is absorbed and in an endothermic process energy is absorbed too.

6 0
3 years ago
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