Write a possible electron configuration for potassium in the excited state

How much water would need to be added to 1092 mL of a 54.7 M NaCl solution to make a 0.25 M solution?

babunello [35]

Answer:

237.8L of water would need to be added.

Explanation:

The first thing to do is to identify that the equation to be used is M1V1=M2V2. (This equation works because it turns everything into moles which can then be compared).

Then figure out what information you have and what is being found. In this case:

M1 = 54.7 M

V1 = 1092 mL = 1.092 L

M2 = 0.25 M

V2 = unknown

Then solve the equation for whatever you are trying to find.

M1V1=M2V2

V2=M1V1/M2

Now you need to plug everything in.

V2=(54.7M*1.091L)/0.25M

V2=238.93L

That means that the solution needs a volume of 238.7L to gain a molarity of 0.25M but the starting solution already had a volume of 1.092 L meaning that to find the amount of solvent that needs to be added you just subtract the starting volume by the volume that the solution needs to be.

238.93L - 1.091L = 237.8L

Therefore the answer is that 237.8L needs to be added to a 1.092L 54.7M NaCl solution to make the concentration 0.25M.

I hope this helps. Let me know if anything is unclear.

8 0

3 years ago

In an experiment, 200.00 grams of Al2O3 was decomposed producing 77.18 grams of Al. Calculate the percent yield for this experim

lakkis [162]

Answer:

approximately 106 grams

Explanation:

The molecular mass of Aluminum Oxide is approximately 102 grams per mole.

2×Al=

2×27=54

3×O=3×16=48

54+48=102

200

102

=1.96 moles

There are two atoms of Aluminum in one mole and each atom has a mass of approximately 27 grams so

Mass of Al = 1.96×2×27=106 grams

Explanation:

6 0

3 years ago

Read 2 more answers

I need this all by today because I’m get 0 and my science teacher is being a

Shalnov [3]

The cnidarias life cycle has 2 life cycles polyp and medusa

7 0

4 years ago

Read 2 more answers

A balloon filled with helium has a volume of 4.5 × 103 L at 25°C. What volume will the balloon occupy at 50°C if the pressure su

Tom [10]

Answer:

$V_2 = 4.87 * 10^3$

Explanation:

This question is an illustration of ideal Gas Law;

The given parameters are as follows;

Initial Temperature = 25C

Initial Volume = 4.5 * 10³L

Required

Calculate the volume when temperature is 50C

NB: Pressure remains constant;

Ideal Gas Law states that;

$PV = nRT$

The question states that the pressure is constant; this implies that the constant in the above formula are P, R and n

Divide both sides by PT

$\frac{PV}{PT} = \frac{nRT}{PT}$

$\frac{V}{T} = \frac{nR}{P}$

Represent $\frac{nR}{P}$ with k

$\frac{V}{T} = k$

$k = \frac{V_1}{T_1} = \frac{V_2}{T_2}$

At this point, we can solve for the required parameter using the following;

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$

Where V1 and V2 represent the initial & final volume and T1 and T2 represent the initial and final temperature;

From the given parameters;

V1 = 4.5 * 10³L

T1 = 25C

T2 = 50C

Convert temperatures to degree kelvin

V1 = 4.5 * 10³L

T1 = 25 +273 = 298K

T2 = 50 + 273 = 323K

Substitute values for V1, T1 and T2 in $\frac{V_1}{T_1} = \frac{V_2}{T_2}$

$\frac{4.5 * 10^3}{298} = \frac{V_2}{323}$

Multiply both sides by 323

$323 * \frac{4.5 * 10^3}{298} = \frac{V_2}{323} * 323$

$323 * \frac{4.5 * 10^3}{298} = V_2$

$V_2 = 323 * \frac{4.5 * 10^3}{298}$

$V_2 = \frac{323 * 4.5 * 10^3}{298}$

$V_2 = \frac{1453.5 * 10^3}{298}$

$V_2 = 4.87 * 10^3$

Hence, the final volume at 50C is $V_2 = 4.87 * 10^3$

7 0

3 years ago

How much does calcite coast

Nikitich [7]

Answer:

It depends on what kind and how much. Some are about $100, and others are $1,000.

5 0

4 years ago