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2 years ago

40 g of CaCO3 is how many moles of CaCO3? 10 moles 0.4 moles 40 moles 100 moles

Chemistry

1 answer:

AleksandrR [38]2 years ago

8 0

Answer:

0.4 moles

Explanation:

To convert between moles and grams you need the molar mass of the compound. The molar mass of of CaCO3 is 100.09g/mol. You use that as the unit converter.

40gCaCO3* 1mol CaCO3/100.09gCaCO3 = 0.399640 mol CaCO3

This rounds to 0.4 moles CaCO3

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-- A scientist combines 14 grams of potassium nitrate with an unknown amount of

igor_vitrenko [27]

Answer:

The unknown amount of potassium chloride is 13.6 grams.

Explanation:

The reaction of 14 grams of KNO₃ with KCl produces a total mass of 27.6 grams of the products.

The law of conservation of mass tells us that the total mass of the reactants must be the same that the total mass of the products. So, we can find the mass of KCl as follows:

$m_{r} = m_{p}$

Where r is for reactants and p is for products

$m_{KNO_{3}} + m_{KCl} = m_{p}$

$14 g + m_{KCl} = 27.6 g$

$m_{KCl} = 27.6 g - 14 g = 13.6 g$

Therefore, the unknown amount of potassium chloride is 13.6 grams.

I hope it helps you!

5 0

3 years ago

If the ligand has a negative charge at a particular location, what would happen if you tried to put electrons from the metal nea

slega [8]

Answer:

The two would end up repelling each other very strongly and more energy would ultimately be required to keep the metal-ligand system in place

Explanation:

A complex is made up a central metal atom or ion and ligands. Ligands are lewis bases and they possess lone pairs of electrons. A complex is formed when electrons are donated from ligand species to metals.

However, if the ligand has a negative charge at a particular location and we try to put electrons from the metal near the electrons from the ligand, the two would end up repelling each other very strongly and more energy would ultimately be required to keep the metal-ligand system in place.

8 0

3 years ago

An acid solution is 0.100 M in HCl and 0.210 M in H2SO4. What volume of a 0.150 M solution of KOH must be added to 500.0 mL of t

Dmitry_Shevchenko [17]

Answer:

When we add 0.33L of KOH only the HCl solution will be neutralized. When we add 1.4 L of KOH, all of the acid (the HCl and H2SO4 solution) will be neutralized.

Explanation:

Step 1: Data given

The solution has 0.100 M HCl and 0.210 M H2SO4

Molarity KOH = 0.150 M

Volume of acid solution = 500 mL = 0.5 L

Step 2: Calculate moles of HCl

Moles HCl = Molarity HCl * volume

Moles HCl = 0.100 M * 0.5 L

Moles HCl = 0.05 moles

Step 3: Calculate moles of H2SO4

Moles H2SO4 = 0.210 M * 0.5 L

Moles H2SO4 = 0.105 moles

Step 4: The balanced neutralization reaction of KOH with HCl can be written as:

KOH + HCl → KCl + H2O

The mole ratio is 1:1

This means to neutralize 0.05 moles HCl, we need 0.05 moles KOH

Step 5: The balanced neutralization reaction of KOH with H2SO4 can be written as:

2KOH + H2SO4 → K2SO4 + 2H2O

The mole ratio KOH: H2SO4 is 2:1

This means to neutralize 0.105 moles H2SO4 we need 0.210 moles KOH

Step 6: Calculate volume of KOH needed to neutralize the solution

To neutralize the HCl solution: 0.05 moles / 0.150 M = 0.33 L KOH needed

To neutralize the H2SO4 solution: 0.210 moles / 0.150 M = 1.4 L KOH needed

When we add 0.33L of KOH the HCl solution will be neutralized. When we add 1.4 L of KOH the HCl and H2SO4 solution will be neutralized.

4 0

2 years ago

Help with chemistry, 50 points

expeople1 [14]

Answer:

Thank you

I didn't wait because there is someone who will take it and there is someone who answer a wrong answer

Explanation:

4 0

2 years ago

Read 2 more answers

Problem PageQuestion Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with o

zlopas [31]

Answer: The amount of acetylene gas needed in the reaction is 2.00 moles

Explanation:

We are given:

Moles of water = 2.00 moles

The chemical equation for the reaction of acetylene with oxygen gas follows:

$2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O$

By Stoichiometry of the reaction:

2 moles of water is produced by 2 moles of acetylene gas

So, 2.00 moles of water will be produced by = $\frac{2}{2}\times 2.00=2.00mol$ of acetylene gas

Hence, the amount of acetylene gas needed in the reaction is 2.00 moles

8 0

2 years ago