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3 years ago

5

What is the electron configuration of the element in period 2 that has 5 valence electrons (valence electrons are the electrons

in the outermost shell) *

1 answer:

balu736 [363]3 years ago

8 0

Answer:

Nitrogen

Explanation:

N= is the symbol

electronic configuration

1s2 2s2 2p3

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PLEASE HELP!!!!

coldgirl [10]

Answer:

1.2 liters.

Explanation:

Focus on the 4th digit: that's the ones column. The 3rd digit is the decimal place, just be sure to round up.

7 0

3 years ago

A flask with a volume of 250.0 mL contains air with a density of 1.164 g/L. What is the mass of the air contained in the flask?

satela [25.4K]

Answer:

convert 250.0 mL in Liters :250. 0 / 1000 = 0,25 LDensity = 1.240 g/LMass

Explanation:

6 0

2 years ago

mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

<u>Answer:</u> The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

<u>Explanation:</u>

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

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

<u>For nitrogen gas:</u>

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

<u>For hydrogen gas:</u>

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

<u>For nitrogen gas:</u>

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

<u>For hydrogen gas:</u>

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

3 years ago

If 38.6 grams of iron react with an excess of bromine gas, what mass of FeBr2 can form?

Yuki888 [10]

Answer:

›› FeBr2 molecular weight. Molar mass of FeBr2 = 215.653 g/mol. This compound is also known as Iron(II) Bromide. Convert grams FeBr2 to moles or moles FeBr2 to grams. Molecular weight calculation: 55.845 + 79.904*2 ›› Percent composition by element

Explanation:

5 0

2 years ago

A 35.0-ml sample of 0.20 m lioh is titrated with 0.25 m hcl. What is the ph of the solution after 23.0 ml of hcl have been added

kiruha [24]

<h3><u>Answer;</u></h3>

pH = 12.33

<h3><u>Explanation;</u></h3>

The equation of reaction is :

LiOH(aq) + HCl(aq) --> LiCl(aq) + H2O(l)

Reactants left after the titrant is added;

Total Moles LiOH;

= 0.035L LiOH × (0.2moles/L)

= 0.007moles of LiOH

Moles of HCl;

= 0.023L HCl × (0.25moles/L)

= 0.00575moles HCl is the limiting reagent

Reacting amount of moles of LiOH;

= 0.0575 moles HCl *(1mole LiOH/1moles HCl)

=0.00575 moles LiOH (reacted)

Moles of LiOH left;

= 0.007moles total - 0.00575moles that react

= .00125 moles of LiOH (left)

LiOH is a strong base, which means that it ionizes completely.

0.00125moles LiOH *(moles/0.058L) = 0.02155M of LiOH

LiOH(aq) --> Li+(aq) + OH-(aq)

[LiOH] = [OH-] = 0.02155 M

pOH = -log[OH-]

pOH = -log(0.02155)

pOH= 1.67

pH = 14 - pOH

pH = 14 - 1.67

pH = 12.33

7 0

3 years ago