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Kaylis [27]
3 years ago
6

How many grams of O2 are required to produce 5 mol CO2

Chemistry
1 answer:
maxonik [38]3 years ago
7 0
I did the math i got 220 grams
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If a solution is given to you, how will you determine whether it is acid,base or salt​
12345 [234]

Answer:

If your lab has litmus paper, you can use it to determine your solution's pH. When you place a drop of a solution on the litmus paper, the paper changes color based on the pH of the solution. Once the color changes, you can compare it to the color chart on the paper's package to find the pH.

Explanation:

A solution's pH will be a number between 0 and 14. A solution with a pH of 7 is classified as neutral. If the pH is lower than 7, the solution is acidic. When pH is higher than 7, the solution is basic. These numbers describe the concentration of hydrogen ions in the solution and increase on a negative logarithmic scale.

For example, If Solution A has a pH of 3 and Solution B has a pH of 1, then Solution B has 100 times as many hydrogen ions than A and is therefore 100 times more acidic.

5 0
3 years ago
Monique hears on the weather report that cold, dry air from Canada is going to hit warm, moist air from the Gulf of Mexico. She
lara [203]

Answer:

She should suspect a Tornado.

Explanation:

Tornadoes form when warm humid air collides with cold dry air. The denser the cold air is pushed over warm air, causing an updraft. As soon as it reaches the ground a tornado is formed.

4 0
3 years ago
Which is the product of that reaction
ki77a [65]

Answer:

B

Explanation:

7 0
3 years ago
Can anybody do this?
Alenkinab [10]

Answer:this should be c

Explanation:

4 0
2 years ago
What pressure is exerted by 932.3 g of CH4 in a 0.560 L steel container at 136.2 K?
Roman55 [17]
Molar mass CH4 = 16.0 g/mol

* number of moles:

932.3 / 16 => 58.26875 moles

T = 136.2 K

V = 0.560 L

P = ?

R = 0.082

Use the clapeyron equation:

P x V = n x R x T

P x 0.560 =  58.26875 x 0.082 x 136.2

P x 0.560 = 650.76

P = 650.76 / 0.560

P = 1162.07 atm
5 0
3 years ago
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