Which of the following best describes the force causing a rocket to lift off?

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibriu

tatyana61 [14]

Answer:

[H2] = 0.012 M

[N2] = 0.019 M

[H2O] = 0.057 M

Explanation:

The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.

2NO(g) + 2H2(g) ⇒ N2(g) + 2H2O(g)

i mol 0.10 0.050 0.10

c mol -0.038 -0.038 +0019 +0.038

e mol 0.062 0.012 00.019 0.057

Since the volume of the vessel is 1.0 L, the concentrations in molarity are:

[NO] = 0.062 M

[H2] = 0.012 M

[N2] = 0.019 M

[H2O] = 0.057 M

5 0

2 years ago

When chlorine and magnesium react chemically, how many chlorine atoms will combine with one magnesium atom?

Lesechka [4]

Magnesium is divalent. This means that one magnesium atom needs to lose 2 electrons in order to become stable.

Chlorine, on the other hand, is monovalent. This means that one chlorine atom needs to gain one electron in order to become stable.

Based on this, one magnesium atom will combine with two chlorine atoms, where the magnesium loses two electrons, one for each chlorine.

The formula of the compound formed is: MgCl2

8 0

2 years ago

How many ounces are in 0.442 L? Use dimensional analysis to

vovangra [49]

Answer:

14.945798 in us fluid ounce

Explanation:

3 0

3 years ago

Using the following reaction:

Romashka-Z-Leto [24]

Answer:

Mn is the oxidizing agent.

N is the reducing agent.

Explanation:

Hello!

In this case, according to the undergoing chemical reaction, it is seen that the manganese in KMnO4 has an oxidation state of 7+, in MnSO4 of 2+ and nitrogen in KNO2 is 3+ and in KNO3 is 5+; thus we have the following half-reactions:

$Mn^{7+}+5e^-\rightarrow Mn^{2+}\\\\N^{3+}\rightarrow N^{5+}+2e^-$

Thus, since manganese is undergoing a decrease in the oxidation state, we infer it is the oxidizing agent whereas nitrogen, undergoing an increase in the oxidation state is the reducing agent.

Best regards!

5 0

2 years ago

onsider the reversible dissolution of lead(II) chloride. P b C l 2 ( s ) − ⇀ ↽ − P b 2 + ( a q ) + 2 C l − ( a q ) PbClX2(s)↽−−⇀

Sveta_85 [38]

Answer:

9.34x10^-4

Explanation:

Step 1:

The balanced equation for the reaction.

PbCl2( s ) <=> Pb^2+(aq) + 2Cl^−(aq)

Step 2:

Data obtained from the question:

Mass of PbCl2 = 0.2393 g

Volume = 50mL

concentration of Pb^2+, [Pb^2+] = 0.0159 M

Concentration of Cl^-, [Cl^-] = 0.0318 M

Equilibrium constant, Kc =?

Step 3:

Determination of the number of mole PbCl2.

The number of mole of PbCl2 can be obtained as follow:

Molar Mass of PbCl2 = 207 + (35.5x2) = 278g/mol

Mass of PbCl2 = 0.2393 g

Number of mole =Mass /Molar Mass

Number of mole of PbCl2 = 0.2393/278 = 8.61x10^-4 mole

Step 4:

Determination of Molarity of PbCl2.

At this stage we shall obtain the molarity of PbCl2. This is shown below:

Mole of PbCl2 = 8.61x10^-4 mole

Volume = 50mL = 50/1000 = 0.05L

Molarity of PbCl2 =?

Molarity = mole /Volume

Molarity of PbCl2 = 8.61x10^-4/0.05

Molarity of PbCl2 = 0.01722 M

Step 5:

Determination of the equilibrium constant Kc.

PbCl2( s ) <=> Pb^2+(aq) + 2Cl^−(aq)

The equilibrium constant Kc for the equation above is given by:

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

[Pb^2+] = 0.0159 M

[Cl^-] = 0.0318 M

[PbCl2] = 0.01722 M

Kc =?

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

Kc = 0.0159 x (0.0318)^2/ 0.01722

Kc = 9.34x10^-4

5 0

3 years ago