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Alexus [3.1K]
3 years ago
9

Can you plz answer this question: The ocean supply about 90 percent of the ————— water that goes inti the water cycle

Chemistry
1 answer:
vovikov84 [41]3 years ago
4 0
B evaporated not a or c
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Rust is formed by the interaction of iron and oxygen. When rust formed on an iron nail with a mass of 100g,the mass of the ruste
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43g id say. Due to conservation of mass that is the only one that makes even slight sense. If you had more air react then the nail would heavier
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Write the chemical formula for Potassium Nitride
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Answer:

KNO lower case 2

Explanation:

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3 years ago
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A mixture of 1.20 mols He, 2.40 mols Ne, 4.80 mols Kr, and 0.60 mols Ar has a total pressure of 600.0 mm Hg. What is the partial
Juliette [100K]

Answer: The partial pressure of the Kr is 320 mm Hg.

Explanation:

According to Raoult's Law , the partial pressure of each component in the solution is equal to the total pressure multiplied by its mole fraction. It is mathematically expressed as

p_A=x_A\times P_{total}

where, p_A = partial pressure of component A

x_A = mole fraction of A

P_{total} = total pressure

mole fraction of Krypton = \frac{\text {Moles of Kr}}{\text {total moles}}=\frac{4.80}{1.20+2.40+4.80+0.60}=0.53

p_{Kr}=0.53\times 600mmHg=320mmHg

Thus partial pressure of the Kr is 320 mm Hg

8 0
3 years ago
According to Boyle’s Law, if temperature and number of gas particles are held constant, what happens to the pressure of a system
larisa86 [58]
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3 years ago
Titration Volume & Concentration
polet [3.4K]

Answer:

18,1 mL of a 0,304M HCl solution.

Explanation:

The neutralization reaction of Ba(OH)₂ with HCl is:

2 HCl + Ba(OH)₂ → BaCl₂ + 2 H₂O

The moles of 17,1 mL≡0,0171L of a 0,161M Ba(OH)₂ solution are:

0,0171L*\frac{0,161moles}{L} = 2,7531x10⁻³moles of Ba(OH)₂

By the neutralization reaction you can see that 2 moles of HCl reacts with 1 mole of Ba(OH)₂. For a complete reaction of 2,7531x10⁻³moles of Ba(OH)₂ you need:

2,7531x10^{-3}molBa(OH)_{2}*\frac{2molHCl}{1molBa(OH)_{2}} = 5,5062x10⁻³moles of HCl.

The volume of a 0,304M HCl solution for a complete neutralization is:

5,5062x10^{-3}molHCl*\frac{1L}{0,304mol} = 0,0181L≡18,1mL

I hope it helps!

4 0
3 years ago
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