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2 years ago

What could you do if you wanted to slow down the rate of a particular reaction?

Chemistry

1 answer:

hjlf2 years ago

5 0

Answer:

catalyst

Explanation:

this help toalter or speed up the rate of reaction could be the most to slow it

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true or false: 1. An object can have only one type of energy at a time 2. If an object has energy, it must be moving. 3. All ene

gregori [183]

True is the correct answer

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2 years ago

Calculate 5+7*3* Your answer:

Mrac [35]

I got the answer 26
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2 years ago

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A fossil was analyzed and determined to have a carbon-14 level that is 70 % that of living organisms. The half-life of C-14 is 5

Citrus2011 [14]

Answer: 2948

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

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

$k=\frac{0.69}{t_{\frac{1}{2}}}=\frac{0.693}{5730}=1.21\times 10^{-4}years^{-1}$

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 = $1.21\times 10^{-4}years^{-1}$

t = age of sample = ?

a = let initial amount of the reactant = 100

a - x = amount left after decay process = $\frac{70}{100}\times 100=70$

$t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{100}{70}$

$t=2948years$

Thus the fossil is 2948 years old.

5 0

2 years ago

A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling

aniked [119]

Answer: The mass of glucose in final solution is 0.420 grams

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL$

Putting values in above equation, we get:

$0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M$

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g$

Hence, the mass of glucose in final solution is 0.420 grams

3 0

3 years ago

Calculate the freezing point of the solution containing 0.105 m k2s. calculate the boiling point of the solution above. -g

Licemer1 [7]

Quantity of K2S m = 0.105 m
Number of ions i = 2(K) + 1(S) = 3
Freezing point depression constant of water Kf = 1.86
delta T = i x m x Kf = 3 x 0.105 x 1.86 = 0.586
Freezing point = 0 - 0.586 = 0.586 C
Boiling point constant of water Kb = 0.512
delta T = i x m x Kb = 3 x 0.105 x 0.512 = 0.161
Boiling point = 100 + 0.161 = 100.161 C

5 0

3 years ago

Read 2 more answers