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Julli [10]
3 years ago
9

What happens to the rate of reaction when you decrease the temperature of a chemical reaction?

Chemistry
2 answers:
Neporo4naja [7]3 years ago
4 0
Depending on what chemicals you use it slows the reaction
Wittaler [7]3 years ago
3 0
Usually it slows the reaction down. This is because of a Kinetic Molecular Theory, which is the idea that molecules need to collide to react. Increasing temperature means they are more likely to collide in a shorter amount of time.
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PLEASE HELP!!!!
shtirl [24]
1) d

2) b because the independent variable is the thing you change/control in an experiment

3) c because the dependent variable is the thing being measured in an experiment

4)hmm it might be d, as c and a are both correct as different sized feeders would make it an unfair test and different types of food would as well

5) c

6) a

7) b obviously because if he activated them at different times then the ones activated last would have an advantage

6 0
3 years ago
What happens to the number of valence electrons as you move from element 1 to element 18 on the periodic table?
BARSIC [14]

Answer:

B

Explanation:

the group number is=valence electrons. element 1 is in group 1 element 18 is in group 8. 1<8

6 0
3 years ago
Read 2 more answers
Science question down below
aliya0001 [1]
Option 1/A (It is the first one)
6 0
3 years ago
To what volume would you need to dilute 200 mL of a 5.85M solution of Ca(OH)2 to make it a 1.95M solution?
Blizzard [7]

Answer: 600 mL

Explanation:

Given that;

M₁ = 5.85 m

M₂ = 1.95 m

V₁ = 200 mL

V₂ = ?

Now from the dilution law;

M₁V₁ = M₂V₂

so we substitute

5.85 × 200 = 1.95 × V₂

1170 = 1.95V₂

V₂ = 1170 / 1.95

V₂ = 600 mL

Therefore final volume is 600 mL

8 0
3 years ago
ubstance A undergoes a first order reaction A®B with a half-life, t½, of 20 min at 25 °C. If the initial concentration of A in a
Stells [14]

Answer : The concentration of A after 80 min is, 0.100 M

Explanation :

Half-life = 20 min

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{20\text{ min}}

k=3.465\times 10^{-2}\text{ min}^{-1}

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 3.465\times 10^{-2}\text{ min}^{-1}

t = time passed by the sample  = 80 min

a = initial amount of the reactant  = 1.6 M

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

80=\frac{2.303}{3.465\times 10^{-2}}\log\frac{1.6}{a-x}

a-x=0.100M

Therefore, the concentration of A after 80 min is, 0.100 M

3 0
3 years ago
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