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2 years ago

Which of the following will affect the rate of a chemical reaction?

Chemistry

2 answers:

dimulka [17.4K]2 years ago

7 0

Answer:

Solute concentration will afect the rate of a chemical reaction, because you must work with molarity

Explanation:

I think that solute mass may be it can affect the rate of reaction, if you have more mass in a solute, you will also have more moles.

If you want to know more, you have to consider temperature in the reaction and the presence of catalysts. They all, affect reactions.

s344n2d4d5 [400]2 years ago

7 0

Answer:

Solute Concentration

Explanation:

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Force = 2.32 N<br> Acceleration = 0.19 m/s2<br> Mass= ?

Mkey [24]

mass = 12.2 kg

Explanation:

To find the mass we use the following formula (Newton's Second Law of Motion):

force = mass × acceleration

mass = force / acceleration

mass = 2.32 N / 0.19 m/s²

mass = 12.2 kg

Learn more about:

Newton's Second Law of Motion

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3 0

2 years ago

Calculate the molality of isoborneol in the product if, a) the melting point of pure camphor is 179°C and the melting point take

tresset_1 [31]

Answer:

The molality of isoborneol in camphor is 0.53 mol/kg.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = $T_f$ = 165°C

Depression in freezing point = $\Delta T_f=?$

$\Delta T_f=T- T_f=179^oC-165^oC=14^oC$

Depression in freezing point is also given by formula:

$\Delta T_f=i\times K_f\times m$

$K_f$ = The freezing point depression constant

m = molality of the sample

i = van't Hoff factor

We have: $K_f$ = 40°C kg/mol

i = 1 ( organic compounds)

$\Delta T_f=14^oC$

$14^oC=1\times 40^oC kg/mol\times m$

$m=\frac{14^oC}{1\times 40^oC kg/mol}=0.35 mol/kg$

The molality of isoborneol in camphor is 0.53 mol/kg.

8 0

2 years ago

Someone help pls thank you i will give brainliest

Lelu [443]

Answer: See attached image

Explanation:

6 0

1 year ago

Read 2 more answers

What two types of particles exist within an atomic nucleus?

MissTica

Protons and Neutrons, the electrons circle around the nucleus.
Hope this Helps!

6 0

2 years ago

You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (p K a = 4.20 ) and 0.200 M sodium benzoate

bekas [8.4K]

Answer : The volume of sodium benzoate and benzoic acid solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.

Explanation :

Let the volume of sodium benzoate (salt) be, x

So, the volume of benzoic acid (acid) will be, (100 - x)

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

Now put all the given values in this expression, we get:

$4.00=4.20+\log \left(\frac{(\frac{0.200x}{100})}{(\frac{0.100(100-x)}{100})}\right)$

x = 29.0

The volume of sodium benzoate = x = 29.0 mL

The volume of benzoic acid (acid) = (100 - x) = (100 - 29.0) = 71 mL

Thus, the volume of sodium benzoate and benzoic acid solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.

5 0

2 years ago