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Anuta_ua [19.1K]
3 years ago
8

2. At STP, a sample of gas occupies 24.5 mL. Calculate the volume of this gas at a

Chemistry
1 answer:
Law Incorporation [45]3 years ago
6 0

Answer:

11.7 ml = V

Explanation: Here

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A mixture of propane and butane is burned with pure oxygen. The combustion products contain 46.7 mole% H2O. After all the water
wlad13 [49]

Answer:

28%

Explanation:

Basically, all o did was write the equations, balance it and solve for them. Also, at the place I stared, I used simultaneous equation to solve it. Multiplying by 8 and also 3.

It's a pretty straightforward question.

At the final step that's missing, I Did

(y)C3H8 = 2.8 / ( 2.8 + 7.1)

(y)C3H8 = 0.28

5 0
3 years ago
Plz answer!!! Due in 10 minutes!!! Will give BRAINLIEST and extra points to first correct answer!!!
Ugo [173]

Answer:

B

Explanation:

u do the math and you will get the answer

8 0
3 years ago
Read 2 more answers
Which element decreases its oxidation number in this reaction? bicl2 + na2so4 → 2nacl + biso4?
kifflom [539]
The balanced reaction is as follows;
BiCl₂ + Na₂SO₄ --> 2NaCl + BiSO₄
this is a double displacement reaction 
the oxidation number of Bi is +2 in both BiCl₂ and BiSO₄
oxidation number of Cl is -1 in both BiCl₂ and NaCl 
oxidation number of Na is +1 in both Na₂SO₄ and NaCl
oxidation numbers of elements in SO₄²⁻ remains the same in both compounds.Therefore the oxidation state in any of the elements in the reaction doesn't change. Neither of the elements show an increase or decrease in the oxidation numbers .
Answer for this question is no element decreases its oxidation number.

5 0
3 years ago
Read 2 more answers
Identify the element using the Bohr model below.​
Maurinko [17]
Nitrogen I believe . I need 20 characters.
3 0
3 years ago
the hemiacetal below is treated with 18o-labeled methanol (ch3o*h) and acid. where will the label appear in the products?
ioda

The 18o-labeled methanol (CH3O*H) will appear in the products side at position b.

<h3>Position of 18o-labeled methanol in the products</h3>

The 18O label will appear at position b in the product as indicated in the image.

This methoxy group in the product formed in position b comes from the 18O-labeled methanol (CH3OH).

While the oxygens at positions a and c in the product come from the unlabeled hemiacetal.

Thus, the 18o-labeled methanol (CH3O*H) will appear in the products side at position b.

Learn more about methanol here: brainly.com/question/17048792
#SPJ11

8 0
2 years ago
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