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2 years ago

Given the following reaction:

Chemistry

1 answer:

Wewaii [24]2 years ago

3 0

Answer:

Approx. 4⋅g.

Explanation:

Moles of sulfuric acid =10.0⋅g98.08⋅g⋅mol−1=0.102⋅mol.

Now we have the molar quantity of sulfuric acid that react; we also have the stoichiometric equation that shows the molar equivalence of sulfuric acid, and lithium hydroxide.

Given the stoichiometry,

mass of water =0.102⋅mol×2×18.01.g.mol−1=??⋅g.

Why did I multiply the mass in this equation by 2? Am I pulling your leg?

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Pa help po science po

mamaluj [8]

Answer:

1. molecule, cell, tissue, organ, organ system, organism, population, community, ecosystem, biosphere

2. A

3. D

4. C

5. C

6. C

7. C

8. B

9. Sorry I don't know :(

10. B

11. A

12. C

13. Sorry :((

14. C

15. a

16. d

Sorry I don't know the rest.. I hope this helps though!!

4 0

1 year ago

A student predicts that a solution of ethanol (C2H5OH) and water will have a lower density at room temperature than that of pure

ohaa [14]

Answer:

It would probably be, something that can take up moisture to test it.

Explanation:

(to see if it can evaporate)

4 0

1 year ago

Read 2 more answers

Find the number of moles of water that can be formed if you have 154 mol of hydrogen gas and 72 mol of oxygen gas.

9966 [12]

Answer:

Explanation:

H2O needs 2 hydrogens and you have 154 so 154/2 is 77

thats the most you can do

7 0

2 years ago

During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce

swat32

Answer: 16.32 g of $O_2$ as excess reagent are left.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol$

$\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol$

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

According to stoichiometry :

2 moles of $SO_2$ require = 1 mole of $O_2$

Thus 0.34 moles of $SO_2$ will require= $\frac{1}{2}\times 0.34=0.17moles$ of $O_2$

Thus $SO_2$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess reagent.

Moles of $O_2$ left = (0.68-0.17) mol = 0.51 mol

Mass of $O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g$

Thus 16.32 g of $O_2$ as excess reagent are left.

3 0

2 years ago

A gas at 400 k is placed in a 4.0 l moveable piston. The gas is cooled to 252 k. What is the new volume of the piston?

katrin [286]

Answer:

2.52L

Explanation:

Given parameters:

T₁ = 400K

V₁ = 4L

T₂ = 252K

unknown

V₂ = ?

Solution:

To solve this problem, we are going to apply charle's law. The law states that the volume of a fixed mass of gas is directly proportional to temperature provided pressure is constant.

Mathematically,

$\frac{V_{1} }{T_{1} } = \frac{V_{2} }{T_{2} }$

Substitute and solve for V₂

$\frac{4}{400} = \frac{V_{2} }{252}$

V₂ = 2.52L

6 0

2 years ago