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3 years ago

which of the following is a balanced chemical equation for the reaction of magnesium with nitrogen gas to form magnesium nitride

1 answer:

ioda3 years ago

8 0

So the unbalanced equation would be Mg + N^2 --> Mg^3N^2

Which means the balanced equation would be 3Mg + N^2 --> Mg^3N^2

This is balance the equation out since you now has 3 magnesium and 2 nitrogen on the left side, and 3 magnesium on 2 nitrogen on the right. Double check my work though, it's been awhile.

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A sample of nitrogen is initially at a pressure of 1.7 kPa, a temperature of -10 C and a volume of 7.5 m3. Then the volume is de

zhannawk [14.2K]

Answer:

$\boxed{\text{2.6 kPa}}$

Explanation:

To solve this problem, we can use the Combined Gas Laws:

$\dfrac{p_{1}V_{1} }{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}$

Data:

p₁ = 1.7 kPa; V₁ = 7.5 m³; T₁ = -10 °C

p₂ = ?; V₂ = 3.8 m³; T₂ = 200 K

Calculations:

(a) Convert temperature to kelvins

T₁ = (-10 + 273.15) K = 263.15 K

(b) Calculate the pressure

$\begin{array}{rcl}\dfrac{1.7 \times 7.5 }{263.15} & = & \dfrac{p_{2} \times 3.8}{200}\\\\0.0485 & = & 0.0190p_{2}\\p_{2} & = & \textbf{2.6 kPa}\\\end{array}\\\text{The new pressure of the gas is \boxed{\textbf{2.6 kPa}}}$

7 0

2 years ago

What best describes how the behavior of rocks changes when they become deeply buried and placed under high levels of heat and pr

Debora [2.8K]

Answer:

They become ductile and deform plastically

Explanation:

When rocks are buried by the materials up to a greater depth, then the confining pressure increases significantly. This results in the ductile behavior of the rocks at such depth. These rocks are present in the ductile region where the depth is about more than 20 to 30 km. Here the rocks are subjected to extremely high pressure and temperature conditions, which favors the transformation of rocks into more higher-grade metamorphic rocks. It is also enhanced due to the geothermal gradient.

Under such high pressure and temperature, the rocks show the behavior of plasticity, where the rocks undergo bending, buckling as well as they tend to flow, and there occurs low strain rate, resulting in the permanent deformation of rocks.

Thus, the rocks become ductile and deform plastically at such conditions.

5 0

3 years ago

What is the final concentration of cl- ion when 250 ml of 0.20 m cacl2 solution is mixed with 250 ml of 0.40 m kcl solution? (as

serious [3.7K]

CaCl2 and KCl are both salts which dissociate in water when dissolved. Assuming that the dissolution of the two salts are 100 percent, the half reactions are:

KCl ---> K+ + Cl-

Therefore the total Cl- ion concentration would be coming from both salts. First, we calculate the Cl- from each salt by using stoichiometric ratio:

Cl- from CaCl2 = (0.2 moles CaCl2/ L) (0.25 L) (2 moles Cl / 1 mole CaCl2)

Cl- from CaCl2 = 0.1 moles

Cl- from KCl = (0.4 moles KCl/ L) (0.25 L) (1 mole Cl / 1 mole KCl)

Cl- from KCl = 0.1 moles

Therefore the final concentration of Cl- in the solution mixture is:

Cl- = (0.1 moles + 0.1 moles) / (0.25 L + 0.25 L)

Cl- = 0.2 moles / 0.5 moles

<span>Cl- = 0.4 moles (ANSWER)</span>

6 0

3 years ago

Which of the following is a testable hypothesis

natka813 [3]

Answer:

Explanation:

8 0

3 years ago

Which solute would be more effective at lowering the freezing point of water: MgCl2 and KNO3? Explain.

Phantasy [73]

Answer:

AlCl₃.

Explanation:

Adding solute to water causes depression of the boiling point.

The depression in freezing point (ΔTf) can be calculated using the relation:

ΔTf = i.Kf.m,

where, ΔTf is the depression in freezing point.

i is the van 't Hoff factor.

van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kf is the molal depression constant of water.

m is the molality of the solution (m = 1.0 m, for all solutions).

(1) NaCl:

i for NaCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (NaCl) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(2) MgCl₂:

i for MgCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

∴ ΔTb for (MgCl₂) = i.Kb.m = (3)(Kf)(1.0 m) = 3(Kf).

(3) NaCl:

i for KBr = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (KBr) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(4) AlCl₃:

i for AlCl₃ = no. of particles produced when the substance is dissolved/no. of original particle = 4/1 = 4.

∴ ΔTb for (CoCl₃) = i.Kb.m = (4)(Kf)(1.0 m) = 4(Kf).

So, the ionic compound will lower the freezing point the most is: AlCl₃

4 0

2 years ago

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