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A chemist prepares a solution of copper(II) fluoride by measuring out of copper(II) fluoride into a volumetric flask and filling

Simora [160]

The question is incomplete, here is the complete question.

A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.

Answer: The concentration of copper fluoride in the solution is $4.90\times 10^{-3}mol/L$

Explanation:

To calculate the molarity of solute, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of copper (II) fluoride = 0.0498 g

Molar mass of copper (II) fluoride = 101.54 g/mol

Volume of solution = 100.0 mL

Putting values in above equation, we get:

$\text{Molarity of copper (II) fluoride)=\frac{0.0498\times 1000}{101.54\times 100.0}\\\\\text{Molarity of copper (II) fluoride}=4.90\times 10^{-3}mol/L$

Hence, the concentration of copper fluoride in the solution is $4.90\times 10^{-3}mol/L$

4 0

3 years ago

Lisa needs to buy a helium tank to fill up all the balloons. If the helium must be compressed from an original volume of 27.0L a

algol13

Answer:

$P_2=135.73psi$

Explanation:

Hello,

In this case we use the Boyle's law which allows us to understand the volume-pressure behavior as an inversely proportional relationship:

$V_1P_1=V_2P_2$

Whereas we solve for $P_2$ as the required final pressure:

$P_2=\frac{V_1P_1}{V_2} =\frac{27.0L*14.7psi}{3.00L}\\ \\P_2=135.73psi$

Best regards.

7 0

3 years ago

Read 2 more answers

Identify the correct coefificients to balance it -C3H8+O2 to -CO2 +-H2O

Iteru [2.4K]

Answer:

${\rm 1\; C_3H_8} + {\rm 5\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}$ .

Explanation:

${\rm ?\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}$ .

Among the four species in this reaction, $\rm C_3H_8$ is species with the largest number of atoms per molecule. Assume that the coefficient of this compound is $1$ .

${\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}$ .

Number of atoms on the left-hand side of the reaction:

$\rm C$ : $1 \times 3 = 3$ .
$\rm H$ : $1 \times 8 = 8$ .
$\rm O$ : not found yet.

By the conservation of atoms, the number of atoms on the right-hand side of the reaction should match those on the left-hand side. In this reaction, $\rm CO_2$ is the only product with carbon atoms, whereas $\rm H_2O$ is the only product with hydrogen atoms. These $3$ carbon atoms and $8$ hydrogen atoms would correspond to:

$3 / 1 = 3$ $\rm CO_2$ molecules, and
$8 / 2 = 4$ $\rm H_2O$ molecules.

${\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}$ .

Number of atoms on the right-hand side of the reaction:

$\rm C$ : $3 \times 1 = 3$ .
$\rm H$ : $4 \times 2 = 8$ .
$\rm O$ : $3 \times 2 + 4 \times 1 = 10$ .

The number of $\rm O$ atoms on the left-hand side should match those on the right-hand side. In this reaction, $\rm O_2$ is the only reactant with $\rm O\!$ atoms. These $10$ $\rm \! O$ atoms would correspond to:

$5$ $\rm O_2$ molecules.

${\rm 1\; C_3H_8} + {\rm 5\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}$ .

5 0

2 years ago

Consider this reaction: 6 CO2 + 6 H2O + light equation C6H12O6 + 6 O2 If there were 2.38 x 102 g of H2O, 18.6 moles of CO2, and

alisha [4.7K]

H₂O would be the limiting reactant.

Balanced chemical equation:

6CO₂ + 6H₂O + light equation → C₆H₁₂O₆ + 6O₂

The amount of product that can be created is constrained by the reactant that is consumed first in a chemical reaction, commonly referred to as the limiting reactant (or limiting reagent).

Given

No. of moles of CO₂ = 18.6

Mass of H₂O = 2.38 × 10² g = 238g

No. of moles of H₂O = Given mass/ Molar mass

= 238 / 18 = 13.22 moles

Moles of H₂O = 13.22

According to the balanced chemical equation

6 moles of CO₂ react with 6 moles of H₂O

So the reactant that has less number of moles will be consumed first.

As the No. of moles of H₂O < No. of moles of CO₂

So, H₂O is the limiting reactant with 13.22 moles.

Hence, H₂O would be the limiting reactant.

Learn more about limiting reactant here brainly.com/question/14222359

#SPJ1

7 0

2 years ago

An orbital is the space occupied by a pair of electrons. true or false??

Anvisha [2.4K]

Answer:

True

Explanation:

An orbital is is the space occupied by a pair of electrons. The maximum number of electrons in an orbital is 2.

The maximum number of electrons in in the orbitals are two.

For s-sublevel with one orbital we have two electrons

p-sublevel with three orbitals we have six electrons

d - sublevel with five orbitals we have ten electrons

f - sublevel with seven orbitals we have fourteen electrons

Each orbital can take a maximum of two electrons.

7 0

3 years ago