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Eddi Din [679]
2 years ago
10

In the synthesis reaction, the white powder produced is magnesium oxide (mgo, mm = 40.3 g/mol). how many moles of magnesium oxid

e were formed (number of moles = mass (g) / molar mass (g/mol))? choose the closest answer.
a.0.21 moles

b.0.41 moles

c.0.11 moles

d.0.31 moles submit for grading save and finish later
Chemistry
1 answer:
gulaghasi [49]2 years ago
3 0
Number of moles = mass of product / molecular mass
=mass of product (MgO) / 40.3

Since the mass of MgO is not given in the question, the correct answer choice cannot be given.  However, proceeding witht eh above formula will enable you to find the correct number of moles given the mass of MgO.
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The concentration of ozone in a sample of air that has a partial pressure of O3 of 0.33 torr and a total pressure of air of 695
Goryan [66]

Answer:

0.047 %

Explanation:

Step 1: Given data

  • Partial pressure of ozone (pO₃): 0.33 torr
  • Total pressure of air (P): 695 torr

Step 2: Calculate the %v/v of ozone in the air

Air is a mixture of gases. We can find the %v/v of ozone (a component) in the air (mixture) using the following expression.

<em>%v/v = pO₃/P × 100%</em>

%v/v = 0.33 torr/695 torr × 100%

%v/v = 0.047 %

8 0
2 years ago
Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −
kicyunya [14]

Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

Mathematically;

ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)  ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca)  = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207  KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

For the last reaction;

ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH  of the respective molecules given above,

ΔH  = -1270 + (-393.5) - (-1207)

ΔH  = -456.5 KJ

8 0
2 years ago
A 360mg sample of aspirin, C9H8O4, (molar mass 180g), is dissolved in enough water to produce 200mL of solution. What is the mol
notsponge [240]

Answer:

Molarity is 0.04M

Explanation:

First of all, let's determinate the moles of aspirin in that sample

Mass / Molar mass = Moles

360 mg = 0.360 g

0.360 g / 180 g/m = 0.002 moles

This moles that are included in 200 mL of solution, are also in 50 mL.

So molarity is mol/L

50 mL = 0.05 L

0.002 m / 0.05 L = 0.04M

6 0
2 years ago
At 800 K, the equilibrium constant, Kp, for the following reaction is 3.2 × 10–7. 2 H2S(g) ⇌ 2 H2(g) + S2(g) A reaction vessel a
djverab [1.8K]

Answer:

0.008945 atm

Explanation:

In the reaction:

2H2S(g) ⇌ 2 H2(g) + S2(g)

Kp is defined as:

Kp = P_{H_{2}}^2P_{S_{2}} / P_{H_{2}S}^2

<em>Where P is the pressure of each compound in equilibrium.</em>

If initial pressure of H2S is 3.00atm, concentrations in equilibrium are:

H2S = 3.00 atm - 2X

H2 = 2X

S2: = X

Replacing:

3.2x10^{-7} = (2X)^2X / (3-2X)^2

3.2x10^{-7} = 4X^3 / 9- 6X+4X^2

0 = 4X³ - 1.28x10⁻⁶X² + 1.92x10⁻⁶X - 2.88x10⁻⁶

Solving for X:

X = 0.008945 atm

As in equilibrium, pressure of S2 is X, <em>pressure is 0.008945 atm</em>

7 0
2 years ago
Write down the dissolution equation for rubidium chromate dissolving in water. (Chromate is a polyatomic ion with the formula Cr
ycow [4]

Answer:

Four moles of the cation

Explanation:

2Rb2CrO4(s)<--------> 4Rb^+(aq) + 2CrO4^2-(aq)

Now looking at the reaction equation, it can be seen that one mole of rubidium chromate contains two moles of rubidium ions and one mole of chromate ions.

The dissolution of two moles of rubidium chromate should then yield four moles of rubidium ions and two moles of chromate ions since the ratio of ions present is 2:1.

This explains the reaction equation written above for the dissolution of two moles of rubidium chromate as shown.

5 0
2 years ago
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