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3 years ago

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An empty 3.00 l bottle weighs 1.70 kg. filled with a certain wine, it weighs 4.72 kg. the wine contains 11.5% ethyl alcohol, c2h

6o, by mass. how many grams of ethyl alcohol are there in 250.0 ml of this wine?

1 answer:

postnew [5]3 years ago

4 0

<span>28.94 g of alcohol. First, calculate the mass of 3.00 l of wine. 4.72 kg = 1.70 kg = 3.02 kg Now calculate the mass of 250.0 ml of the wine. Using ratios will work for this (3020 g)/(3000 ml) = (X g)/(250 ml) Multiply both sides by 250 ml (3020 g * 250 ml) / (3000 ml) = X (755000 ml g) / (3000 ml) = X 251.667 g = X So 250.0 ml of wine has a mass of 251.667 g Now multiply by 0.115 to get the mass of alcohol 251.667 g * 0.115 = 28.94 g</span>

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What is the isotope notation for an ion of silver-109 with a charge of positive 1

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3 years ago

A gas has a pressure of 1.34 atm when the temperature is 237K. The gas is then heated until the temperature measures 312K. What

alexandr1967 [171]

The answer for the following question is answered below.

<em><u>Therefore the new pressure of the gas is 1.76 atm.</u></em>

Explanation:

Given:

Initial pressure of the gas = 1.34 atm

Initial temperature of the gas = 273 K

final temperature of the gas = 312 K

To solve:

Final temperature of the gas

We know;

From the ideal gas equation

P × V = n × R × T

So;

from the above equation we can say that

<em>P ∝ T</em>

$\frac{P}{T}$ = constant

$\frac{P_{1} }{P_{2} }$ = $\frac{T_{1} }{T_{2} }$

Where;

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<em><u>Therefore the new pressure of the gas is 1.76 atm.</u></em>

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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

3 years ago