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Step2247 [10]
3 years ago
9

How many liters of a 6.0M solution Fe(SCN)2 are needed to prepare 2.0L of a 0.75M solution?

Chemistry
2 answers:
scoray [572]3 years ago
8 0

Answer:

We need 0.25 L of the 6.0 M solution

Explanation:

Step 1: Data given

Molarity of the Fe(SCN)2 solution = 6.0 M

Volume of the 0.75 M solution = 2.0 L

Step 2: Calculate volume

C1*V1 = C2*V2

⇒ with C1 = the initial concentration = 6.0 M

⇒ with V1 = the initial volume = TO BE DETERMINED

⇒ with C2 = the new concentration = 0.75 M

⇒ with V2 = the new volume = 2.0 L

6.0M * V1 = 0.75M * 2.0L

V1 = (0.75 M * 2.0L) / 6.0 M

V1 = 0.25 L

We need 0.25 L of the 6.0 M solution

lubasha [3.4K]3 years ago
4 0

Answer:

0.25L

Explanation:

Using the dilution formula

C1V1=C2V2

C1=6M

V1?

C2=0.75M

V2=2.0L

V1= C2V2/C1

V1=0.75*2.0/6

V1=0.25L

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3 years ago
Searches related to If 0.75 grams of iron (Fe) react according to the following reaction, how many grams of copper (Cu) will be
jolli1 [7]

Answer:

0.83 g

Explanation:

Step 1: Write the balanced equation

Fe + CuSO₄ ⇒ Cu + FeSO₄

Step 2: Calculate the moles corresponding to 0.75 g of Fe

The molar mass of Fe is 55.85 g/mol.

0.75g \times \frac{1mol}{55.85g} = 0.013 mol

Step 3: Calculate the moles of Cu produced from 0.013 moles of Fe

The molar ratio of Fe to Cu is 1:1. The moles of Cu produced are 1/1 × 0.013 mol = 0.013 mol.

Step 4: Calculate the mass corresponding to 0.013 moles of Cu

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4 0
4 years ago
Read 2 more answers
A 6.35 l sample of carbon monoxide is collected at 55.0◦c and 0.892 atm. What volume will the gas occupy at 1.05 atm and 59.0◦c?
Sonja [21]

Answer : The final volume of gas will be, 5.46 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 0.892 atm

P_2 = final pressure of gas = 1.05 atm

V_1 = initial volume of gas = 6.35 L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 55.0^oC=273+55.0=328K

T_2 = final temperature of gas = 59.0^oC=273+59.0=332K

Now put all the given values in the above equation, we get:

\frac{0.892atm\times 6.35L}{328K}=\frac{1.05atm\times V_2}{332K}

V_2=5.46L

Thus, the final volume of gas will be, 5.46 L

7 0
3 years ago
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