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Triss [41]
3 years ago
8

Convert 3.6 x 10^2 atm to torr

Chemistry
1 answer:
bezimeni [28]3 years ago
3 0

Answer:

3.6 x (10^2) atm =  273 600 torr

Explanation:

Hope that helps

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The Earth's mantle is
EleoNora [17]

Answer:

C. cooler than both the crust and the core

Explanation:

It is observed that at the mantle, temperatures range from estimatedly 200 °C (392 °F) around the upper boundary with the crust to approximately 4,000 °C (7,230 °F) at the core-mantle boundary.

So we can say the mantle is cooler than both the crust and the core.

5 0
3 years ago
Which of the following is the best definition of convection?
lara [203]
<span>a) movement of heat through space </span>
7 0
3 years ago
Read 2 more answers
Which of these abiotic factors is most likely the reason some plants cannot live very close to the ocean?
Blababa [14]
C.  dim light is <span>the answer
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7 0
2 years ago
Are the following solutions acidic, basic, ot neutral.
Slav-nsk [51]

Answer:

1) Basic

2) Basic

3) Acidic

Explanation:

pH of a solution is a measure of the H3O+ ions in the solution and hence reflects its acidity.

pH = -log[H3O+]

Solutions are classified as acidic, basic or neutral based on the pH range

-pH < 7, acidic

- pH = 7, neutral

- pH > 7, basic

1) [H3O+] = 2.5*10^-9M

pH = -log[H3O+]=-log[2.5*10^{-9}]=8.60

Since pH > 7, solution is basic

2)[OH-] = 1.6*10^-2M

pOH = -log[OH-]=-log[1.6*10^{-2}]=1.80

pH = 14 - pOH = 14 - 1.80 =12.2

Since pH > 7, solution is basic

3) [H3O+] = 7.9*10^-3M

pH = -log[H3O+]=-log[7.9*10^{-3}]=2.10

Since pH < 7, solution is acidic

6 0
3 years ago
2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)a. Determine the volume (mL) of 15.0 M sulfuric acid needed to react with 45.0 g of
Ira Lisetskai [31]

Answer:

a. 167 mL.

b. 39.3 %.

Explanation:

Hello!

In this case, for the undergoing chemical reaction, since 45.0 g of aluminum react, based on the 2:3 mole ratio with sulfuric acid, we can compute the required moles  as shown below:

n_{H_2SO_4}=45.0gAl*\frac{1molAl}{27.0gAl} *\frac{3molH_2SO_4}{2molAl} =2.50molH_2SO_4

Next, since the molarity of a solution is computed based on the moles and volume (M=n/V), we can compute the required volume of sulfuric acid as shown below:

V=\frac{n}{M}=\frac{2.50mol}{15.0mol/L}=0.167L

That in mL is 167 mL.

Moreover, for the percent yield, we compute the grams of aluminum sulfate that are produced based on the required 2.50 moles of sulfuric acid:

m_{Al_2(SO_4)_3}=2.50molH_2SO_4*\frac{1molAl_2(SO_4)_3}{3molH_2SO_4}*\frac{342.15gAl_2(SO_4)_3}{1molAl_2(SO_4)_3} \\\\m_{Al_2(SO_4)_3}=285.13gAl_2(SO_4)_3

Therefore the percent yield is:

Y=\frac{112g}{285.13g}*100\%\\\\Y=39.3\%

Best regards!

6 0
2 years ago
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