The Kw of water varies with temperature. Calculate the pH of water at 46⁰C with a Kw = 1.219 x 10-14. Show all calculations leading to an answer.

Answers

Answer 1
Answer: Kw is equal to the concentration of the H+ and the OH-. In pure water, the concetration of H+ and the OH- is equal. In computing for the pH of water, we tke the negative log of the concentration of H+. To solve the concentration of H+ we take the square root of the Kw (Since Kw=[H+]*[H+] since [H+]=[OH-]) 
H+ is equal to 1.104*10^-7
negative log of the H+ conc. or the pH is 6.96 (pH = -log(1.104*10^-7))

The answer is valid and is close to the neutral pH which is 7. 

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Which two gases make up most of the atmosphere?

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nitrogen wich comprises 78% of the atmosphere and oxygen wich accounts for 21%

Write these chemicals in order starting with the one that turns into gas the easiest. Naphtha, diesel, bitumen, petrol

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What is the percent yield of ferrous sulfide if the actual yield is 220.0 g and the theoretical yield is 275.6 g

Answers

To get the percent yield, we will use this formula: ((Actual Yield)/(Theoretical Yield)) * 100% Values given: actual yield is 220.0 g theoretical yield is 275.6 g Now, let us substitute the values given. (220.0 grams)/(275.6 grams) = 0.7983 Then, to get the percentage, multiply the quotient by 100. 0.7983 (100) = 79.83% Among the choices, the most plausible answer is 79.8%
If the actual yield of ferrous sulfide is 220.0 g and its theoretical yield is 275.6 g, the percent yield of ferrous sulfide is 79.8%. Make a proportion. 220.0 g is x percent. 275.6 g is 100%. 220.0 g : x = 275.6 g : 100%. After crossing the products: x = 220.0 g * 100% : 275.6 g = 79.8%

a student used 10 mL water instead of 30 mL for extraction of salt from mixture. How may this change the percentage of NaCl extracted?

Answers

It will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.

If it is known that solubility of NaCl is 360 g/L, let's find out how many NaCl is in 30 mL of water:

360 g : 1 L = x g : 30 mL

Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = x g : 30 mL

Now, crossing the products:
· 1,000 mL = 360 g · 30 mL
x · 1,000 mL = 10,800 g mL
x = 10,800 g ÷ 1,000 
x = 10.8 g

So, from 30 mL mixture, 10.8 g of NaCl could be extracted.

Let's calculate the same for 10 mL water instead of 30 mL.

360 g : 1 L = x g : 10 mL

Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = x g : 10 mL

Now, crossing the products:
· 1,000 mL = 360 g · 10 mL
x · 1,000 mL = 3,600 g mL
x = 3,600 g ÷ 1,000 
x = 3.6 g

So, from 10 mL mixture, 3.6 g of NaCl could be extracted.

Now, let's compare:
If from 30 mL mixture, 10.8 g of NaCl could be extracted and from 10 mL mixture, 3.6 g of NaCl could be extracted, the ratio is:
3.6/10.8 = 1/3

Therefore, it will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water. 

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Answers

the force that accelerates objects towards Earth is gravity.

If 455-mL of 6.0 m HNO3 is used to make a 2.5L dilution, what is the molarity of the dilution

Answers

2.5 L * 1000 = 2500 mL

C₁ * V₁ = C₂ * V₂

6.0 * 455 = C₂ * 2500

2730 = C₂ * 2500

C₂ = 2730 / 2500

C₂ = 1.092 M

hope this helps!