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kakasveta [241]
2 years ago
10

Name some Changes in matter...

Chemistry
2 answers:
frutty [35]2 years ago
5 0

Answer:

Solid, Liquid, Gas.

Explanation:

Quick gg search.

Bess [88]2 years ago
5 0

Answer:

melting

freezing

sublimation

deposition

condensation

vaporization

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How to balance equations
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start the balancing by writing down how many atoms there are per element. we’ll use this as an example:

C3H8 + O2 --> H2O + CO2

C = 3 C = 1

H = 8 H = 2

O = 2 O = 3

balance the carbon first, as it is easiest to do. add a coefficient to the single carbon atom on the right of the equation to balance it with the 3 carbon atoms on the left of the equation:

C3H8 + O2 --> H2O + (3)CO2

now there are 3 carbon atoms on each side. however, when you do this, you multiply the amount of oxygen atoms you had. therefore, now, there are 6 carbon atoms in 3CO2, plus that other oxygen atom in H2O. you now have 7 O atoms instead of 3.

C = 3 C = 3

H = 8 H = 2

O = 2 O = 7

now let’s move on to the hydrogen atoms.

C3H8 + O2 --> H2O + 3CO2

you have 8 hydrogen atoms on the left side, and 2 on the right. in order to balance them, you have to multiply the right side’s hydrogen atoms by 4. 4(2) = 8.

C3H8 + O2 --> (4)H2O + 3CO2

now both hydrogen and carbon atoms are balanced. same amount on both sides. however, your oxygen atoms have changed due to the multiplying (right side). you now have 10 of them.

C = 3 C = 3

H = 8 H = 8

O = 2 O = 10

now we balance the oxygen atoms. multiply the left side of the equation’s oxygen atoms by 5. 5(2) = 10

C3H8 + (5)O2 --> 4H2O + 3CO2

the chemical equation is all balanced. basically, just multiply with numbers until it equals the same amount on both sides.

C = 3 C = 3

H = 8 H = 8

O = 10 O = 10

4 0
3 years ago
How many grams of sodium chloride are required to make a 1.0 L solution with a concentration of 3.5 M? 205 grams 0.205 grams 3.5
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<span>A 1 molar solution is the molecular weight in grams in 1 litre of water, so a 3.5 molar solution would be 58.44g multiplied by 3.5, which is 204.54g in 1L.</span>
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./.....//...//./..//..//././../
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Answer:no question is written

Explanation:

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