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kirza4 [7]
3 years ago
5

For a gas to become a liquid it’s particles must A. Speed up B. Slow down C. Stop

Chemistry
1 answer:
madam [21]3 years ago
3 0
Slow down because liquids have less movement than gases
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What should iron oxide be categorized
11111nata11111 [884]

Answer:

What are you doing stepbro?

Explanation:

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3 years ago
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The arrangement of things in the order that they occurred.
RoseWind [281]
The answer is B: chronological order
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Explain how you were able to use your knowledge of how different types of blood react with anti A, anti B, and anti Rh antibodie
dybincka [34]

Depending upon the clumping reaction with anti A , anti B and anti Rh antibodies the blood types are determined.

Explanation:

Agglutination (clumping) will occur when blood that contains the particular antigen is mixed with the particular antibody.

A+ have Agglutination with Anti-A ,Anti-Rh and No agglutination with Anti-B.

A- have Agglutination with Anti-A and No agglutination with Anti-B and Anti-Rh.

B+ have Agglutination with Anti-B Anti-Rh and No agglutination with Anti-A.

B- have Agglutination with Anti-B and No agglutination with Anti-B and Anti-Rh.

Rh+ have Agglutination with Anti-A and Anti-Rh and No agglutination with Anti-B.

Rh- have No Agglutination with Anti-A and Anti-B and Anti-Rh.

3 0
3 years ago
8.) If 396 g of Carbon Dioxide (CO2) are produced, what mass of Oxygen<br> (02) reacted? *
Brut [27]

Answer:

264g

Explanation:

C + O2 -> CO2

_g + _g -> 396g

396÷3=132

C (132g) + O2 (264g) -> CO2 (396g)

7 0
3 years ago
Use the following balanced reaction to solve:
Naily [24]

Answer:  60.7 g of PH_3 will be formed.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar volume}}    

\text{Moles of} H_2=\frac{60L}{22.4L}=2.68moles

The balanced chemical reaction is

P_4(s)+6H_2(g)\rightarrow 4PH_3(g)

H_2 is the limiting reagent as it limits the formation of product and P_4 is the excess reagent.

According to stoichiometry :

6 moles of H_2 produce = 4 moles of PH_3

Thus 2.68 moles of H_2 will produce=\frac{4}{6}\times 2.68=1.79moles  of PH_3

Mass of PH_3=moles\times {\text {Molar mass}}=1.79moles\times 33.9g/mol=60.7g

Thus 60.7 g of PH_3 will be formed by reactiong 60 L of hydrogen gas with an excess of P_4

3 0
3 years ago
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