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3 years ago

What’s the molar mass of magnesium oxalate

Chemistry

1 answer:

maria [59]3 years ago

8 0

Hey there!

Magnesium oxalate: MgC₂O₄

Mg: 1 x 24.305 = 24.305

C: 2 x 12.01 = 24.02

O: 4 x 16 = 64

----------------------------------

112.325 g/mol

The molar mass of magnesium oxalate is 112.325 g/mol.

Hope this helps!

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If a 66.5 g sample of ammonium nitrate (NH4NO3) is dissolved in enough water to make 315 mL of solution, what will be the molari

11111nata11111 [884]

Answer:

2.64 M

Explanation:

To find the molarity, you need to (1) convert grams to moles (via molar mass), then (2) convert mL to L, and then (3) calculate the molarity (via molarity ratio). The final answer should have 3 sig figs to match the sigs figs of the given values.

(Step 1)

Molar Mass (NH₄NO₃): 2(14.007 g/mol) + 4(1.008 g/mol) + 3(15.998 g/mol)

Molar Mass (NH₄NO₃): 80.04 g/mol

66.5 grams NH₄NO₃ 1 mole
--------------------------------- x ---------------------- = 0.831 moles NH₄NO₃
80.04 grams

(Step 2)

1,000 mL = 1 L

315 mL 1 L
-------------- x ------------------ = 0.315 L
1,000 mL

(Step 3)

Molarity = moles / volume

Molarity = 0.831 moles / 0.315 L

Molarity = 2.64 M

7 0

2 years ago

What mass of carbon dioxide is formed when 1.75 mol of ethane burns completely in oxygen?

dybincka [34]

Answer:

There is a mass of 154 Grams of Carbon Dioxide.

Explanation:

One mole is equal to 6.02 × 10^23 particles.

This means we have 1.05 X 10^24 total particles of Ethane.

Each ethane particle contains 2 carbon atoms.

If every particle of ethane is burned, we will end up with 2.10 x 10^24 molecules of Carbon Dioxide (Particles of Methane x 2, since each Methane particle contains 2 carbon atoms)

Carbon Dioxide has a molar mass of 44.01 g/mol

So if we take our amount of Carbon Dioxide molecules and divide it by 1 mole, ((2.10 x 10^24)/(6.02 x 10^23) = 3.49) we find that we have 3.49 moles of Carbon Dioxide.

Now all we need to do is multiply our moles of carbon dioxide(3.49) by it's molar mass(44.01) while accounting for significant digits.

What you should end up with is 154 Grams of Carbon Dioxide.

Hope this helps (And more importantly I hope I didn't make any errors in my math lol)

As a side note this is all assuming that this takes place at STP conditions.

3 0

3 years ago

A gas occupies 3.5L at 2.5 kPa pressure. What is the volume at 100 mmHg at the same temperature? Be sure to

nataly862011 [7]

Answer:

V₂ = 0.656 L

Explanation:

Given data:

Initial volume = 3.5 L

Initial pressure = 2.5 KPa

Final volume = ?

Final pressure = 100 mmHg (100/7.501=13.33 KPa)

Solution:

The given problem will be solved through the Boyle's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume

Now we will put the values in formula,

P₁V₁ = P₂V₂

2.5 KPa × 3.5 L = 13.33 KPa × V₂

V₂ = 8.75 KPa. L/13.33 KPa

V₂ = 0.656 L

7 0

3 years ago

What molarity of nitric acid (HNO3) was used if 2.00 L must be used to prepare 4.5 L of a 0.25 M HNO3 solution?

Ymorist [56]

The molarity of (HNO₃) that was used if 2.00 L must be used to prepare 4.5 L of a 0.25M HNO₃ solution is 0.563 M

<u><em>calculation</em></u>

This is calculated usind M₁V₁=M₂V₂ formula

where,

M₁( molarity ₁) = ?

V₁( volume ₁) = 2.00 L

M₁ (molarity ₂) = 0.25M

V₂( volume₂) = 4.5 L

make M₁ the subject of the formula by diving both side of the formula by V₁

M₁ is therefore = M₂V₂/V₁

M₁ =[ (0.25 M x 4.5 L) / 2.00 L ] =0.563 M

5 0

3 years ago

1.65g of zinc is used to make 8g of zinc iodide. How much iodine is required for this reaction?

creativ13 [48]

Answer:

6.45 g of iodine, I₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

Zn + I₂ —> ZnI₂

Next, we shall determine the mass of Zn and I₂ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of Zn = 65 g/mol

Mass of Zn from the balanced equation = 1 × 65 = 65 g

Molar mass of I₂ = 127 × 2 = 254 g/mol

Mass of I₂ from the balanced equation = 1 × 254 = 254 g

SUMMARY:

From the balanced equation above,

65 g of Zn reacted with 254g of I₂.

Finally, we shall determine the mass of f I₂ needed to react with 1.65 g of Zn. This can be obtained as follow:

From the balanced equation above,

65 g of Zn reacted with 254g of I₂.

Therefore, 1.65 g of Zn will react with = (1.65 × 254)/65 = 6.45 g of I₂.

Thus, 6.45 g of iodine, I₂ is needed for the reaction.

4 0

3 years ago