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3 years ago

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Two students were in Chemistry Class completing an assigned lab experiment. The two students were to mix 5.34g CaCl2 with CuSO4

to make a 6.3L solution
. Complete and balance the reaction with the correct Products:
CaCl2 + CuSO4  ______ + ______

1 answer:

skelet666 [1.2K]3 years ago

3 0

Answer:

CaCl₂ + CuSO₄ → CaSO₄ + CuCl₂.

Explanation:

It is a replacement reaction, that cations and anions are replaced with each other.

So, it gives CaSO₄ and CuCl₂.

To balance the reaction, we apply the law of conversation of mass, that the no. of atoms in the reactants side is equal to that in the products side.

So, the balanced reaction is:

<em>CaCl₂ + CuSO₄ → CaSO₄ + CuCl₂.</em>

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Fe(s) + CuSO4(aq) <===> Cu(s) + FeSO4(aq)

Ludmilka [50]

Answer : The original concentration of copper (II) sulfate in the sample is, $5.6\times 10^{-1}g/L$

Explanation :

Molar mass of Cu = 63.5 g/mol

First we have to calculate the number of moles of Cu.

Number of moles of Cu = $\frac{\text{Mass of Cu}}{\text{Molar mass of Cu}}=\frac{89\times 10^{-3}g}{63.5g/mol}=1.40\times 10^{-3}mole$

Now we have to calculate the number of moles of $CuSO_4$

Number of moles of Cu = Number of moles of $CuSO_4$

Number of moles of $CuSO_4$ = $1.40\times 10^{-3}mole$

Now we have to calculate the molarity of $CuSO_4$

$\text{Molarity}=\frac{\text{Moles of }CuSO_4\times 1000}{\text{Volume of solution (in mL)}}$

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{1.40\times 10^{-3}mole\times 1000}{400.mL}=0.0035M$

To change mol/L into g/L, we need to multiply it with molar mass of $CuSO_4$

Molar mass of $CuSO_4$ = 159.609 g/mL

Concentration in g/L = $0.0035M\times 159.609g/mol=0.5586g/L\approx 5.6\times 10^{-1}g/L$

Thus, the original concentration of copper (II) sulfate in the sample is, $5.6\times 10^{-1}g/L$

3 0

3 years ago

A flask contains a gas mixture of hydrogen, nitrogen and methane. The total pressure of the mix is 3.0 atm. The partial pressure

ollegr [7]

Answer:

$P_{N_2}=1.0atm$

Explanation:

Hello,

Considering the Dalton's law which states that the total pressure of a gaseous system is defined by the summation of the the partial pressures of the present gases:

$P_T=\Sigma P_i$

For the given system:

$P_T=P_{H_2}+P_{N_2}+P_{CH_4}$

Solving for the partial pressure of nitrogen we obtain:

$P_{N_2}=3.0atm-0.5atm-1.5atm=1.0atm$

Best regards.

3 0

3 years ago

The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.

olchik [2.2K]

Answer:

Approximately $0.291\; \rm M$ (rounded to two significant figures.)

Explanation:

The unit of concentration $\rm M$ is the same as $\rm mol \cdot L^{-1}$ (moles per liter.) On the other hand, the volume of both the $\rm NaOH$ solution and the original $\rm HCl$ solution here are in milliliters. Convert these two volumes to liters:

$V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L$ .
$V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0\times 10^{-3}\; \rm L = 0.0100\; \rm L$ .

Calculate the number of moles of $\rm NaOH$ in that $0.0182\; \rm L$ of $0.160\; \rm M$ solution:

$\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}$ .

$\rm HCl$ reacts with $\rm NaOH$ at a one-to-one ratio:

$\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l)$ .

Coefficient ratio:

$\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1$ .

In other words, one mole of $\rm NaOH$ would neutralize exactly one mole of $\rm HCl$ . In this titration, $0.291\; \rm mol$ of $\rm NaOH\!$ was required. Therefore, the same amount of $\rm HC$ should be present in the original solution:

$\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}$ .

Calculate the concentration of the original $\rm HCl$ solution:

$\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M$ .

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3 years ago

How many orbitals are in the n=3 level

solniwko [45]

The third shell has 3 subshells: the subshell, which has 1 orbital with 2 electrons, the subshell, which has 3 orbitals with 6 electrons, and the subshell, which has 5 orbitals with 10 electrons, for a total of 9 orbitals and 18 electrons.

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The entropy of an exothermic reaction decreases. This reaction will be spontaneous under which of the following temperatures?

kakasveta [241]

Answer:Low temperatures

Explanation:

∆G= ∆H-T∆S

If ∆H is negative (exothermic reaction), then in order to maintain ∆G<0 which is the condition for spontaneity; T must decrease. This is because, decrease in T will keep the difference of ∆H and T∆S at a negative value in order to satisfy the above stated condition for spontaneity.

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3 years ago

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