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irakobra [83]
3 years ago
5

A large gently sloping volcano with a fluid eruption

Chemistry
1 answer:
kozerog [31]3 years ago
6 0
Salutations!

A large gently sloping volcano with a fluid eruption is called shield volcano. The fluid is basaltic lava and the slops are gently sliding.

Hope I helped (:

Have a great day!


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What is the epicenter?
hoa [83]

Answer:

basically the middle of an earthquake

5 0
3 years ago
A sample of Mo(NO3)6 has 2.22 x 10^22 nitrogen atom, how many oxygen atoms does the sample have?
goldfiish [28.3K]

There are 3.98 × 10^23 atoms of oxygen in the sample.

Given that;

1 mole of Mo(NO3)6 contains 6.02 × 10^23 atoms of Nitrogen

x moles of Mo(NO3)6 contains 2.22 x 10^22 atoms of nitrogen

x = 1 mole × 2.22 x 10^22 atoms/6.02 × 10^23 atoms

x = 0.0368 moles

The number of oxygen atoms in the sample is given by; 0.0368 × 6.02 × 10^23 × 18

Therefore, there are 3.98 × 10^23 atoms of oxygen in the sample.

Learn more: brainly.com/question/9743981

4 0
2 years ago
A lab technician mixes a 0.730 M solution of sodium bromide (NaBr) and water. The volume of the solution is 135 milliliters.
Ivenika [448]

Answer:

We need 10.14 grams of sodium bromide to make a 0.730 M solution

Explanation:

Step 1: Data given

Molarity of the sodium bromide (NaBr) = 0.730 M

Volume of the sodium bromide solution = 135 mL = 0.135 L

Molar mass sodium bromide (NaBr) = 102.89 g/mol

Step 2: Calculate moles NaBr

Moles NaBr = Molarity NaBr * volume NaBr

Moles NaBr = 0.730 M * 0.135 L

Moles NaBr = 0.09855 moles

Step 3: Calculate mass of NaBr

Mass NaBr = 0.09855 moles * 102.89 g/mol

Mass NaBr = 10.14 grams

We need 10.14 grams of sodium bromide to make a 0.730 M solution

3 0
3 years ago
Read 2 more answers
Please help..... who is up for a zoom to help me with balancing equations ​
Alex787 [66]

Answer:

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8 0
3 years ago
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Which of the following are true statements about equilibrium systems? For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(
Grace [21]

Answer:

The first, third and fourth statements are correct.

Explanation:

1) For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(s) + CO2(g) adding more CaCO3 will shift the equilibrium to the right.

⇒ Le Chatellier says As the CaCO3 concentration is increased, the system will attempt to undo that concentration change by shifting the balance to the right. <u>This statement is true.</u>

<u />

2) For the following reaction at equilibrium: CaCO3(s)⇌ CaO(s) + CO2(g) increasing the total pressure by adding Ar(g) will shift the equilibrium to the right.

⇒ Le chatellier says that if we increase the pressure, the equilibrium will shift to the side with the least number of particles.

Since the molar densities of CaO and CaCO3 are constant, they don't appear in the equilibrium expression. This is why only changes to the pressure (concentration) of CO2 affect the position of the equilibrium.

If the pressure in the container is increased by adding an inert or non-reacting gas, nothing happens to the amounts of CO2, CaO or CaCO3. The added gas won't affect the partial pressure of CO2. <u>This statement is false. </u>

3)For the following reaction at equilibrium: 2 H2(g) + O2(g) ⇌ 2 H2O(g) the equilibrium will shift to the left if the volume is doubled.

⇒ Le Chatellier says if we increase the pressure, the equilibrium will shift to the side with the most particles.

In this case we have 2 moles of H2 and 1 mole of O2 on the left side and 2 mole of H2O on the right side. This means on the left side are more particles. So the equilibrium will shift to the left, so <u>this statement is true.</u>

4) For the following reaction at equilibrium: H2(g) + F2(g) ⇌ 2HF(g) removing H2 will increase the amount of F2 present once equilibrium is reestablished. Increasing the temperature of an endothermic reaction shifts the equilibrium position to the right.

⇒ Le chatellier says if H2 will be removed (this means the left side will get less particles) so the equilibrium will shift to the left, to increase the amount of F2.

⇒Le chatelier says if we increase the temperature of an exotherm reaction , there will be less energy released. The equilibrium will shift to the side of the reactants (the left side).

If we increase the temperature of an endotherm reaction, the equilibrium will shift to the side of the products (the right side). <u>This statement is true.</u>

4 0
3 years ago
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