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2 years ago

8

Brad and Matt are working in the lab. They noticed that when they mixed two dilute solutions together, the reaction between them

happened very slowly. Which of Matt's suggestions would BEST help to increase the rate of this reaction?

1 answer:

Tju [1.3M]2 years ago

3 0

Answer:

c.) increase the concentration of one of the solutions

Explanation:

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I/3 of 12 = 2/3 of<br>please help at 11 o'clock am it needs to be done

Nostrana [21]

Answer:

6

Explanation:

This question is seeking for an equivalent of one expression. The first expression says 1/3 of 12, which means 1/3 × 12 = 4. To find 2/3 of a number that will give us 4, we say

Since; 1/3 of 12 is 4

Then, 4 = 2/3 of x, where x is the number, which when multiplied by two-third (2/3) will result to 4.

4 = 2/3 × x

4 = 2x/3

Cross multiply

2x = 3 × 4

2x = 12

x = 12/2

x = 6.

This means that; 1/3 of 12 = 2/3 of 6

4 0

2 years ago

What is the m∠L? 45° 54° 40° 52°

Nataly [62]

Answer:

m<L = 45

Explanation:

4 0

2 years ago

Pls help me with this

Zina [86]

Answer:

$[I_2]=[Br]=0.31M$

Explanation:

Hello there!

In this case, according to the given information, it is possible for us to set up the following chemical equation at equilibrium:

$I_2+Br_2\rightleftharpoons 2IBr$

Now, we can set up the equilibrium expression in terms of x (reaction extent) whereas the initial concentration of both iodine and bromine is 0.5mol/0.250L=2.0M:

$K=\frac{[IBr]^2}{[I_2][Br_2]} \\\\1.2x10^2=\frac{(2x)^2}{(2.0-x)^2}$

Thus, we solve for x as show below:

$\sqrt{1.2x10^2} =\sqrt{\frac{(2x)^2}{(2.0-x)^2}} \\\\10.95=\frac{2x}{2.0-x}\\\\21.91-10.95x=2x\\\\21.91=12.95x\\\\x=\frac{21.91}{12.95} \\\\x=1.69M$

Therefore, the concentrations of both bromine and iodine are:

$[I_2]=[Br]=2.0M-1.69M=0.31M$

Regards!

8 0

3 years ago

How many mL of 0.100 M NaOH are needed to neutralize 50.00 mL of a 0.150 M solution of CH3CO2H, a monoprotic acid? How many mL o

Rama09 [41]

Answer:

We need 75 mL of 0.1 M NaOH ( Option C)

Explanation:

<u>Step 1: </u>Data given

Molarity of NaOH solution = 0.100 M

volume of 0.150 M CH3COOH = 50.00 mL = 0.05 L

<u>Step 2:</u> The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

<u>Step 3:</u> Calculate moles of CH3COOH

Moles CH3COOH = Molarity * volume

Moles CH3COOH = 0.150 M * 0.05 L

Moles CH3COOH = 0.0075 moles

<u>Step 4</u>: Calculate moles of NaOH

For 1 mol of CH3COOH we need 1 mol of NaOH

For 0.0075 mol CH3COOH we need 0.0075 mole of NaOH

<u>Step 5:</u> Calculate volume of NaOH

volume = moles / molarity

volume = 0.0075 moles / 0.100 M

Volume = 0.075 L = 75 mL

We need 75 mL of 0.1 M NaOH

7 0

3 years ago

A solution is made by dissolving 42.3g of potassium hydroxide in 329g of water what is the molality of the solution?

xxMikexx [17]

Answer:

2.34 molality

Explanation:

Mole weight of KOH ( using periodic table ...rounded to 3 s.f.) =

56.1 gm/mole

Number of moles in 43.2 g

43.2 g / 56.1 gm / mole =.770 mole

molality = .770 mole / .329 kg = 2.34 m

7 0

1 year ago