Ask question

Ask question

All categories

4 years ago

15

What is the concentration of FeCl3 in a solution prepared by dissolving 10.0 g of FeCl3 in enough water to make 275 mL of soluti

on?What is the concentration of FeCl3 in a solution prepared by dissolving 10.0 g of FeCl3 in enough water to make 275 mL of solution?0.224 M4.46 × 103 M4.46 M2.24 × 10-4 M

1 answer:

puteri [66]4 years ago

7 0

Answer: The concentartion of solution will be 0.224 M

Explanation:

Molarity : It is defined as the number of moles of solute present per liter of the solution.

Formula used :

$Molarity=\frac{n\times 1000}{V_s}$

where,

n= moles of solute = $\frac{\text{Given mass}}{\text{Molar mass}}=\frac{10.0g}{162.2g/mol}=0.0616moles$

${V_s}$ = volume of solution in ml = 275 ml

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{0.0616\times 1000}{275ml}=0.224mole/L$

Therefore, the concentration of solution will be 0.224 M

You might be interested in

Click the DeltaH is an Extensive Property button within the activity, and analyze the relationship between the two reactions tha

DaniilM [7]

Answer:

∴ΔH₂ = - 12,258 KJ

Explanation:

Enthalpy:

Enthalpy is a property of a thermodynamic system. Enthalpy of a system is equal to the sum of internal energy of the system and presser times volume of the system.

The heat absorbes or releases in a closed system is the change of enthalpy of the system.

Given reactions are:

Reaction 1: C₃H₈(g)+5O₂(g)→ 3CO₂(g)+4H₂O, ΔH₁= - 2043 KJ

Reaction 2: 6C₃H₈(g)+30 O₂(g)→ 18 CO₂(g)+24 H₂O, ΔH₂=?

Take a look at reaction 1 and reaction 2, the only difference is that 1 molecule of C₃H₈ is combusted in reaction 1 and 6 molecules of C₃H₈ is combusted in reaction 2.

We can think the reaction 2 as occurring 6 different container and each containers contains 1 molecule of C₃H₈. The enthalpy is an extensive property. Total enthapy of the 6 containers is = 6×(-2043 KJ)

= - 12,258 KJ

∴ΔH₂ = - 12,258 KJ

6 0

4 years ago

Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic

ale4655 [162]

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

$\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9$

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = $164.5\times 10^{-9} g$

$1 ng=10^{-9} g$

Volume of the sample = V = $15.3 cm^3$

Density of the lake water sample ,d= $1.00 g/cm^3$

Mass of sample = M = $d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g$

$ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75$

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in $1 cm^3$ of lake water = $\frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g$

Mass of arsenic in $0.710 km^3$ lake water be m.

$1 km^3=10^{15} cm^3$

Mass of arsenic in $0.710\times 10^{15} cm^3$ lake water :

$m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g$

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

$\frac{2.74}{41.90} days$

Then days required to remove 7,633.66 kg of arsenic from the lake water :

$=7,633.66\times \frac{2.74}{41.90} days=499.19 days$

1 year = 365 days

499.19 days = $\frac{499.19}{365} years = 1.367 years\approx 1.37 years$

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

$\frac{2.74}{41.90} days$

Then days required to remove 7,633.66 kg of arsenic from the lake water :

$=7,633.66\times \frac{2.74}{41.90} days=499.19 days$

1 year = 365 days

499.19 days = $\frac{499.19}{365} years = 1.367 years\approx 1.37 years$

It will take 1.37 years to remove all of the arsenic from the lake.

6 0

3 years ago

Metal in period 6 group 1

Nady [450]

The answer is ceslum

4 0

4 years ago

Balance the following skeleton ionic reaction, and calculate E°cell to decide whether the reaction is spontaneous: Ag(s) + Cu2+

tia_tia [17]

Answer:

2Ag(s) + Cu^2+(aq) ----------> 2Ag^+(aq) + Cu(s)

Explanation:

Ag(s)/Ag^+ (aq) is the anode as shown while Cu^2+(aq)/Cu^2(s) is the cathode.

E°cell= E°cathode -E°anode= 0.34 -0.80= -0.5V

The cell is not spontaneous as written because E°cell is negative. This implies that the electrodes of the cell must be interchanged to make the cell spontaneous.

5 0

4 years ago

Read 2 more answers

Based on the greenhouse effect, if the amount of carbon dioxide in the air decreased, what would happen?

choli [55]

A) the average global temp. Would decrease

7 0

4 years ago