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e-lub [12.9K]
3 years ago
8

Which law is based on the graph that is shown below?

Chemistry
2 answers:
IRINA_888 [86]3 years ago
8 0
I don’t see a graph
lesantik [10]3 years ago
6 0

Answer:

Where's the graph?

Explanation:

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Gerry is looking at an unknown element. He knows it has 1 valence electron and is highly reactive with water. What group is the
SVEN [57.7K]

Answer:

Group 1

because they have just one valance electron, group 1 elements are very reactiv.

4 0
3 years ago
1. Mass is measured with and instrument called a _____
Brilliant_brown [7]
1.triple beam balance
4 0
4 years ago
Pls answer this question and have a AMAGING day!! :))
ladessa [460]

Answer:

B

Explanation:

because it is being cooled down

hoped this helps

3 0
3 years ago
9. Assuming all other conditions are constant, what is the new pressure of a gas if the original pressure is 50 kPa and the Kelv
notsponge [240]

Answer:

Option a (100 kPa) is the appropriate option.

Explanation:

The given value is:

Original pressure,

P₁ = 50 kPa

Let the new pressure be "x".

Now,

⇒  \frac{P1}{T_1} =\frac{P_2}{T_2}

On substituting the values, we get

⇒  \frac{50}{T_1} =\frac{x}{2T_1}

On applying cross-multiplication, we get

⇒  x = 50\times 2

⇒     =100 \ kPa

Thus the answer above is the right one.

8 0
3 years ago
Given the propane molecule C3H8(g)! a) What is the balanced equation for the combustion of propane in O2?(20 pts.) b) What is th
lubasha [3.4K]

Answer:

The answer to your question is below

Explanation:

Propane = C₃H₈

Process

1.- Write the chemical reaction

                     C₃H₈  +  O₂   ⇒   CO₂  +  H₂O

balanced chemical reaction

                      C₃H₈  +  5O₂   ⇒  3CO₂  +  4H₂O

               Reactants      Elements     Products

                      3                    C                  3

                      8                    H                  8

                     10                    O                 10

b) Standard enthalpy

Propane                  -104.7 kJ/mol

Oxygen                         0   kJ/mol

Carbon dioxide      -393.5 kJ/mol

Water                      -241.8 kJ/mol

ΔH° = ∑ΔH° products - ∑H° reactants

ΔH° = 3(-393.5) + 4(-241.8) - [(-104.7) + 0]

-Simplification

ΔH° = -1180.5 kJ/mol - 967.2 kJ/mol + 104.7 kJ/mol

ΔH° = - 2042.5 kJ/mol

This reaction is exothermic because ΔH° is negative.

4 0
3 years ago
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