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s344n2d4d5 [400]
4 years ago
5

You have at your disposal 3 1-pound bags of various pure salts that dissolve readily in water. You can add one of the bags to a

55 gallon barrel of water. Which bag should you use to lower the freezing point by the greatest amount? (1 lb. = 0.454 kg)
Chemistry
1 answer:
Sedaia [141]4 years ago
6 0

Answer:

This question appears incomplete

Explanation:

However, it should be noted that addition of soluble salts generally lowers the freezing point of water hence after the addition, water will no longer freeze at 0°C but lower.

Soluble salts tend to form more ions in water, it is these ions that are responsible for interfering with the hydrogen bonds hence lowering the freezing. Thus, (since each bag are of the same weight) <u>the bag that contains the salt that ionizes more in water will lower the freezing point by the greatest amount</u>.

NOTE: Different weight of the salts could lead to more ions been formed in the water by some salts as against the other.

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P4O10 -&gt; 4P+5O2 How many moles of phosphorus would be produced if 5.3 mol of P4O10 reacted?
viktelen [127]

Answer:

21.2 moles.

Explanation:

Hello!

In this case, for the given chemical reaction, we can see there is a 1:4 mole ratio between tetraphosphorous decaoxide and phosphorous; therefore, the following proportional factor provides the requested moles of phodphorous:

n_P=5.3molP_4O_{10} *\frac{4molP}{1molP_4O_{10}} \\\\n_P=21.2molP

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5 0
3 years ago
92 grams of ethanol,C2H5OH, is dissolved in 1 liter of solution. Determine the molarity (M)
const2013 [10]

Answer:  1.997 M

Explanation:

molarity = moles of solute/liters of solution or M = \frac{mol}{L}

first we have to find our moles of solute (mol), which you can find by dividing the mass of solute by molar mass of solute

mass of solute: 92 g

molar mass of solute: 46.08 g/mol

let's plug it in:

\frac{92 g}{46.08 g/mol} = 1.997 mol

next, we plug it into our original equation:

\frac{1.997mol}{1 L} = 1.997 M

4 0
3 years ago
What is the percent error when a student measures the volume to be 19.3 liters when
katrin2010 [14]

Answer:

The answer is

<h2>13.84 %</h2>

Explanation:

The percentage error of a certain measurement can be found by using the formula

P(\%) =  \frac{error}{actual \:  \: number}  \times 100\% \\

From the question

actual volume = 22.4 L

error = 22.4 - 19.3 = 3.1

The percentage error is

P(\%) =  \frac{3.1}{22.4}  \times 100 \\  = 13.839285714...

We have the final answer as

<h3>13.84 %</h3>

Hope this helps you

7 0
3 years ago
A 3.00-kg block of copper at 23.0°C is dropped into a large vessel of liquid nitrogen at 77.3 K. How many kilograms of nitrogen
hammer [34]

Answer:

1.2584kg of nitrogen boils.

Explanation:

Consider the energy balance for the overall process. There are not heat or work fluxes to the system, so the total energy keeps the same.

For the explanation, the 1 and 2 subscripts will mean initial and final state, and C and N2 superscripts will mean copper and nitrogen respectively; also, liq and vap will mean liquid and vapor phase respectively.

The overall energy balance for the whole system is:

U_1=U_2

The state 1 is just composed by two phases, the solid copper and the liquid nitrogen, so: U_1=U_1^C+U_1^{N_2}

The state 2 is, by the other hand, composed by three phases, solid copper, liquid nitrogen and vapor nitrogen, so:

U_2=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}

So, the overall energy balance is:

U_1^C+U_1^{N_2}=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}

Reorganizing,

U_1^C-U_2^C=U_{2,liq}^{N_2}+U_{2,vap}^{N_2}-U_1^{N_2}

The left part of the equation can be written in terms of the copper Cp because for solids and liquids Cp≅Cv. The right part of the equation is written in terms of masses and specific internal energy:

m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_1^{N_2}u_1^{N_2}

Take in mind that, for the mass balance for nitrogen, m_1^{N_2}=m_{2,liq}^{N_2}+m_{2,vap}^{N_2},

So, let's replace m_1^{N_2} in the energy balance:

m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,liq}^{N_2}u_1^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}

So, as you can see, the term m_{2,liq}^{N_2}u_{2,liq}^{N_2} disappear because u_{2,liq}^{N_2}=u_{1,liq}^{N_2} (The specific energy in the liquid is the same because the temperature does not change).

m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}

m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}(u_{2,vap}^{N_2}-u_1^{N_2})

The difference (u_{2,vap}^{N_2}-u_1^{N_2}) is the latent heat of vaporization because is the specific energy difference between the vapor and the liquid phases, so:

m_{2,vap}^{N_2}=\frac{m_C*Cp*(T_1^C-T_2^C)}{(u_{2,vap}^{N_2}-u_1^{N_2})}

m_{2,vap}^{N_2}=\frac{3kg*0.092\frac{cal}{gC} *(296.15K-77.3K)}{48.0\frac{cal}{g}}\\m_{2,vap}^{N_2}=1.2584kg

3 0
4 years ago
If two pretty beat friends exist would they actually be real or just 2 dogs<br>1.yes<br>2.no
12345 [234]

Answer:

I wouldn’t know I aint never seen two be pretty best friends

Explanation:

4 0
3 years ago
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