Ask question

Ask question

All categories

2 years ago

15

To make a solution for an experiment, Gunther needs to add 40 g of a solute to 100 g of water. When making the solution at room

temperature, he could only add 34 grams before the solute settled out.
What could he do to dissolve the remaining 6 grams of the solute?

Put the solution in an ice bath, dissolve the solute, and let the solution return to room temperature.
Heat the solution, dissolve the solute, and let the solution cool verifying nothing settled out.
Add more water, boil the solution, and dissolve the solute until the some of the water evaporates.
Keep the solution at room temperature, add more water, and dissolve the excess solute.

1 answer:

777dan777 [17]2 years ago

5 0

Answer:

B. Heat the solution, dissolve the solute, and let the solution cool verifying nothing settled out.

Explanation:

You might be interested in

Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit

Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

$0.0811 M=\frac{n}{0.0017 L}$

n = 0.0001378 mol

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

$\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol$ of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

$x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L$

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0

2 years ago

Calculate the number of grams of xenon in 4.658 g of the compound xenon tetrafluoride.

andrezito [222]

Answer:

The mass of xenon in the compound is 2.950 grams

Explanation:

Step 1: Data given

Mass of XeF4 = 4.658 grams

Molar mass of XeF4 = 207.28 g/mol

Step 2: Calculate moles of XeF4

Moles XeF4 = mass XeF4 / molar mass XeF4

Moles XeF4 = 4.658 grams / 207.28 g/mol

Moles XeF4 = 0.02247 moles

Step 3: Calculate moles of xenon

XeF4 → Xe + 4F-

For 1 mol xenon tetrafluoride, we have 1 mol of xenon

For 0.02247 moles XeF4 we have 0.02247 moles Xe

Step 4: Calculate mass of xenon

Mass xenon = moles xenon * molar mass xenon

Mass xenon = 0.02247 moles * 131.29 g/mol

Mass xenon = 2.950 grams

The mass of xenon in the compound is 2.950 grams

5 0

2 years ago

What is I2O8 compound name?

julsineya [31]

Answer:

Formula in Hill system is I2O8. Computing molar mass (molar weight). To calculate molar mass of a chemical compound enter its formula and click 'Compute'. In chemical formula you may use: Any chemical element.

Explanation:

7 0

1 year ago

What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the de

Snezhnost [94]

Here is the full question

Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm until the height of the alcohol column is 10 cm. The two liquids do not mix.

What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the density of glycerin is 1260 kg/m3and the density of alcohol is 790 kg/m3.

Express your answer in two significant figures and include the appropriate units (in cm)

Answer:

ΔH ≅ 3.73 cm

Explanation:

The pressure inside a liquid is known as hydrostatic pressure and which is represent by the formula:

P = ρ × g × h

where;

ρ is the density of the fluid

g is the gravitational constant

h is the height from the surface

From the question above;

For glycerine; we have:

density of glycerine = 1260 kg/m³

gravitational constant = 9.8 m/s²

height = ???

∴

$P_{(g)= 1260kg/m^3}*9.8m/s^2*h_g$ ----- equation (1)

On the other hand for alcohol:

density of alcohol is given as = 790 kg/m³

gravitational constant = 9.8 m/s²

height = 10 cm

∴

$P_{(a)= 790kg/m^3*9.8m/s^2*10$ ----------- equation (2)

if we equate equation 1 and 2 together; we have

$P_{(g)= P_{(a)$

$1260kg/m^3}*9.8m/s^2*h_g = 790kg/m^3*9.8m/s^2*10cm$

Making $h_g$ the subject of the formula, we have :

$h_g= \frac{ 790kg/m^3*9.8m/s^2*10cm}{1260kg/m^3*9.8m/s^2}$

$h_g$ = 6.269 cm

The difference in the height denoted by ΔH can therefore be calculated as:

ΔH $= H_a-H_g$

ΔH $= 10cm - 6.269cm$

ΔH = 3.731 cm

ΔH ≅ 3.73 cm (to two significant figures)

5 0

2 years ago

Carbonic acid, H2CO3, has two acidic hydrogens. A solution containing an unknown concentration of carbonic acid is titrated with

stepan [7]

Answer:

1) Net ionic equation :

$2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)$

2) 0.765 M is the molarity of the carbonic acid solution.

Explanation:

1) In aqueous carbonic acid , carbonate ions and hydrogen ion is present.:

$H_2CO_3(aq)\rightarrow 2H^+(aq)+CO_3^{2-}(aq)$ ..[1]

In aqueous potassium hydroxide , potassium ions and hydroxide ion is present.:

$KOH(aq)\rightarrow K^+(aq)+OH^{-}(aq)$ ..[2]

In aqueous potassium carbonate , potassium ions and carbonate ion is present.:

$K_2CO_3(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)$ ..[3]

$H_2CO_3(aq)+2KOH(aq)\rightarrow K_2CO_3(aq)+2H_2O(l)$

From one:[1] ,[2] and [3]:

$2H^+(aq)+CO_3^{2-}(aq)+2K^+(aq)+2OH^{-}(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)+H_2O(l)$

Cancelling common ions on both sides to get net ionic equation :

$2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)$

2)

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2CO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=2\\M_1=?\\V_1=50.0 mL\\n_2=1\\M_2=3.840 M\\V_2=20.0 mL$

Putting values in above equation, we get:

$M_1=\frac{1\times 3.840 M\times 20.0 mL}{2\times 50.2 mL}=0.765 M$

0.765 M is the molarity of the carbonic acid solution.

6 0

2 years ago