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3 years ago

A solution is made by mixing 34.5 g of sugar with 75.0 g of water. What is the mass percent of sugar in this solution?

Chemistry

1 answer:

lubasha [3.4K]3 years ago

5 0

the mass percent of sugar in this solution is 46%.

Answer:

Solution given:

mass of solute=34.5g

mass of solvent=75g

mass percent= $\frac{mass\:of\:solute}{mass\:of \:solvent}*100\%$

= $\bold{\frac{34.5}{75.0}*100\%=46\%}$

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2 years ago

Provide a balanced molecular equation, total ionic, and net ionic equation for sodium phosphate and zinc acetate.

vivado [14]

Answer: Balanced molecular equation :

$2Na_3PO_4(aq)+3(CH_3COO)_2Zn(aq)\rightarrow 6CH_3COONa(aq)+Zn_3(PO_4)_2(s)$

Total ionic equation:

$6Na^+(aq)+3PO_4^{2-}(aq)+6CH_3COO^-(aq)+3Zn^{2+}(aq)\rightarrow 6CH_3COO^-(aq)+6Na^+(aq)+Zn_3(PO_4)_2(s)$

The net ionic equation:

$2PO_4^{3-}(aq)+3Zn^{2+}(aq)\rightarrow Zn_3(PO_4)_2(s)$

Explanation:

Complete ionic equation : In complete ionic equation, all the substances that are strong electrolyte are present in an aqueous state as ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

When sodium phosphate and zinc acetate then it gives zinc phosphate and sodium acetate as product.

The balanced molecular equation will be,

$2Na_3PO_4(aq)+3(CH_3COO)_2Zn(aq)\rightarrow 6CH_3COONa(aq)+Zn_3(PO_4)_2(s)$

The total ionic equation in separated aqueous solution will be,

$6Na^+(aq)+2PO_4^{3-}(aq)+6CH_3COO^-(aq)+3Zn^{2+}(aq)\rightarrow 6CH_3COO^-(aq)+6Na^+(aq)+Zn_3(PO_4)_2(s)$

In this equation, and are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$2PO_4^{3-}(aq)+3Zn^{2+}(aq)\rightarrow Zn_3(PO_4)_2(s)$

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2 years ago

If Iron (Fe) gained a proton what would it be?

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If iron (Fe) gained a proton it would become cobalt (CO).

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2 years ago

If you combine 230.0 mL 230.0 mL of water at 25.00 ∘ C 25.00 ∘C and 120.0 mL 120.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is t

Thepotemich [5.8K]

Answer: The final temperature of the mixture is 49°C

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

For cold water:

Density of cold water = 1 g/mL

Volume of cold water = 230.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{230.0mL}\\\\\text{Mass of water}=(1g/mL\times 230.0mL)=230g$

For hot water:

Density of hot water = 1 g/mL

Volume of hot water = 120.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{120.0mL}\\\\\text{Mass of water}=(1g/mL\times 120.0mL)=120g$

When hot water is mixed with cold water, the amount of heat released by hot water will be equal to the amount of heat absorbed by cold water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of hot water = 120 g

$m_2$ = mass of cold water = 230 g

$T_{final}$ = final temperature = ?°C

$T_1$ = initial temperature of hot water = 95°C

$T_2$ = initial temperature of cold water = 25°C

c = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$120\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)]$

$T_{final}=49^oC$

Hence, the final temperature of the mixture is 49°C

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