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Snezhnost [94]
3 years ago
13

What happens to the number of valence electrons as you move from left to right in the periodic table?

Chemistry
1 answer:
fomenos3 years ago
6 0
Valence electrons increases
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W(N)={2M(N)/M(NH₄NO₃)}*100%

w(N)={2*14.0 g/mol/80.0 g/mol}*100% = 35%
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When 2.35 g of potassium reacts with excess water, what mass of hydrogen gas is formed?
Butoxors [25]

Answer:

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The number at the end of an isotope's name is the number
Korolek [52]

Answer:

Atomic neutron mass electron number

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7 0
3 years ago
Need help on this question asap pleasee
algol [13]

Answer:

9.0 moles of CaO

Explanation:

We have the reaction equation as follows;

Fe2O3 + Ca3(PO4)2 -------> 2FePO4 + 3CaO

Now we know from the equation that;

1 mole of iron III oxide yields 3 moles of CaO

Therefore;

3 moles of iron III oxide yields 3 * 3/1

= 9.0 moles of CaO

6 0
2 years ago
A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She k
Sergeeva-Olga [200]

<u>Answer:</u> The partial pressure of oxygen gas is 2.76 bar

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant = 0.0831\text{ L bar }mol^{-1}K^{-1}

n = Total number of moles = ?

Putting values in above equation, we get:

5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

p_{A}=p_T\times \chi_{A}         ........(1)

where,

p_A = partial pressure of carbon dioxide = 0.318 bar

p_T = total pressure = 5.57 bar

\chi_A = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571

  • Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, \chi_{N_2}=\frac{0.221}{0.493}=0.448

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar

Hence, the partial pressure of oxygen gas is 2.76 bar

6 0
3 years ago
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