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3 years ago

Which is the formula for magnesium nitride?

Chemistry

2 answers:

horsena [70]3 years ago

7 0

Answer:

the answer is in the picture

Explanation:

Mekhanik [1.2K]3 years ago

5 0

Explanation:

Mg₃N₂ is the formula of magnesium nitride.

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Which substance is an electrolyte <br> 1) CCl4 <br> 2)SiO2<br> 3)C6H12O6<br> 4) H2SO4

Elena L [17]

Electrolyte is any species which when dissolved in solvent particularly water dissociates into cations and anions. Electrolytes are conductors of electricity. In given options;

CCl₄ (Tetrachloromethane) is a covalent compound. And it doesn't dissociate to any cation or anion. So it is not electrolyte.

SiO₂ (Silicon Dioxide) is also covalent in nature and exist in giant framework. It is not electrolyte.

Glucose (C₆H₁₂O₆) is also covalent compound. And doesn't produced any ion in water, hence it is not electrolyte.

H₂SO₄ (Sulfuric acid) is Electrolyte. When it is dissolved in water it produces H⁺ and SO₄²⁻ ions as follow,

H₂SO₄ → 2 H⁺ ₍aq₎ + SO₄²⁻ ₍aq₎

Result:
H₂SO₄ is electrolyte.

5 0

3 years ago

Zinc reacts with lead (II) nitrate to produce zinc nitrate and lead. This reactions is:

Allisa [31]

Answer:

Displacement Reaction

Explanation:

Zinc replaces lead in Lead(ll) nitrate as zinc is more reactive than lead.

3 0

3 years ago

I need help solving this for chemistry. Don’t know where to start:/

Juli2301 [7.4K]

1 electron has charge =1.602* 10⁻¹⁹ C
1 mole of electrons have 1.602* 10⁻¹⁹*6.02*10²³C = 9.64*10⁴ C/1mol

One ion Co²⁺ takes 2e⁻ to become Co⁰.
1 mol of Co²⁺ ions take 2 mole of e⁻ to become Co⁰, so
0.30 mol Co²⁺ ions take mole of 0.60 mol e⁻ to become Co⁰

9.64*10⁴(C/1mol) *0.60 (mol)≈ 5.8 *10⁴ Coulombs.
Correct answer is C

8 0

3 years ago

Is the narrator male or female in the story leaving?

AlladinOne [14]

The narrator is a female

Hope this helps

8 0

3 years ago

A student weighs an empty flask and stopper and finds the mass to be 55.844 g. She then adds about 5 mL of an unknown liquid and

Oduvanchick [21]

Answer :

(a) The pressure of the vapor in the flask in atm is, 0.989 atm

(b) The temperature of the vapor in the flask in Kelvin is, 372.7 K

The volume of the flask in liters is, 0.2481 L

(c) The mass of vapor present in the flask was, 0.257 g

(d) The number of moles of vapor present are 0.00802 mole.

(e) The mass of one mole of vapor is 32.0 g/mole

Explanation : Given,

Mass of empty flask and stopper = 55.844 g

Volume of liquid = 5 mL

Temperature = $99.7^oC$

Mass of flask and condensed vapor = 56.101 g

Volume of flask = 248.1 mL

Barometric pressure in the laboratory = 752 mmHg

(a) First we have to determine the pressure of the vapor in the flask in atm.

Pressure of the vapor in the flask = Barometric pressure in the laboratory = 752 mmHg

Conversion used :

$1atm=760mmHg$

or,

$1mmHg=\frac{1}{760}atm$

As, $1mmHg=\frac{1}{760}atm$

So, $752mmHg=\frac{752mmHg}{1mmHg}\times \frac{1}{760}atm=0.989atm$

Thus, the pressure of the vapor in the flask in atm is, 0.989 atm

(b) Now we have to determine the temperature of the vapor in the flask in Kelvin.

Conversion used :

$K=273+^oC$

As, $K=273+^oC$

So, $K=273+99.7=372.7$

Thus, the temperature of the vapor in the flask in Kelvin is, 372.7 K

Now we have to determine the volume of the flask in liters.

Conversion used :

1 L = 1000 mL

or,

1 mL = 0.001 L

As, 1 mL = 0.001 L

So, 248.1 mL = 248.1 × 0.001 L = 0.2481 L

Thus, the volume of the flask in liters is, 0.2481 L

(c) Now we have to determine the mass of vapor that was present in the flask.

Mass of flask and condensed vapor = 56.101 g

Mass of empty flask and stopper = 55.844 g

Mass of vapor in flask = Mass of flask and condensed vapor - Mass of empty flask and stopper

Mass of vapor in flask = 56.101 g - 55.844 g

Mass of vapor in flask = 0.257 g

Thus, the mass of vapor present in the flask was, 0.257 g

(d) Now we have to determine the number of moles of vapor present.

Using ideal gas equation:

PV = nRT

where,

P = Pressure of vapor = 0.989 atm

V = Volume of vapor = 0.2481 L

n = number of moles of vapor = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of vapor = 372.7 K

Putting values in above equation, we get:

$(0.989atm)\times 0.2481L=n\times (0.0821L.atm/mol.K)\times 372.7K\\\\n=0.00802mole$

Thus, the number of moles of vapor present are 0.00802 mole.

(e) Now we have to determine the mass of one mole of vapor.

$\text{Mass of one mole of vapor}=\frac{\text{Mass of vapor}}{\text{Moles of vapor}}$

$\text{Mass of one mole of vapor}=\frac{0.257g}{0.00802mole}=32.0g/mole$

Thus, the mass of one mole of vapor is 32.0 g/mole

8 0

3 years ago