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horrorfan [7]
3 years ago
6

A chemical process used to produce ethanol as a fuel additive was expected to produce 5,000 kilograms of ethanol based on the am

ounts of starting materials used, but only 4,760 kilograms were produced. What was the percent yield for ethanol in this process?
A)1.09
B)4.80
C)95.2
D)105
Chemistry
1 answer:
yarga [219]3 years ago
7 0

Answer:

C)95.2

Explanation:

Given parameters:

Expected yield  = 5000kg

Experimental  yield  = 4760kg

Unknown:

Percentage yield  = ?

Solution:

The percentage yield of the substance is determined by find the ratio of the true experimental yield to the expect yield and multiply by 100;

  Percentage yield  = \frac{4760}{5000} x 100 = 95.2%

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What is the amount of heat released by 1.00 gram of liquid water at 0°C when it changes to 1.00 gram of ice at 0°C?
QveST [7]

Answer:

334J/g

Explanation:

Data obtained from the question include:

Mass (m) = 1g

Specific heat of Fusion (Hf) = 334 J/g

Heat (Q) =?

Using the equation Q = m·Hf, we can obtain the heat released as follow:

Q = m·Hf

Q = 1 x 334

Q = 334J

Therefore, the amount of heat released is 334J

8 0
4 years ago
1. The solubility of AgNO3 at 20°C is 222.0g AgNO3/100g H2O. What mass of AgNO3 can be dissolved in 250 g of water at 20°C? Reca
neonofarm [45]

Answer :

(1) The mass of silver nitrate is, 555 g

(2) The solubility of the gas will be, 0.433 g/L

<u>Solution for Part 1 :</u>

From the given data we conclude that

In 100 gram of water, the amount of silver nitrate = 222 g

In 250 gram of water, the amount of silver nitrate = \frac{222}{100}\times 250=555g

Therefore, the mass of silver nitrate is, 555 g

<u>Solution for Part 2 :</u>

Formula used : S_1P_1=S_2P_2     (at constant temperature)

where,

S_1 = initial solubility of methane gas = 0.026 g/L

S_2 = final solubility of methane gas

P_1 = initial pressure of methane gas = 1 atm

P_2 = final pressure of methane gas = 0.06 atm

Now put all the given values in the above formula, we get the solubility of methane gas.

(0.026g/L)\times (1atm)=S_2\times (0.06atm)

S_2=0.433g/L

Therefore, the solubility of the gas will be, 0.433 g/L

5 0
3 years ago
If 4.50 g of (nh4)2so4 is dissolved in enough water to form 250. ml of solution, what is the molarity of the solution?
natima [27]
The molarity of a solution is the number of moles of a substance in one liter of that substance. 
The molar mass of ammonium sulfate (NH4)2SO4 is 132.14 grams/mole
Calculate the moles of ammonium sulfate:
(4.50 grams)/(132.14 grams/mole) = 0.0341 moles of ammonium sulfate
convert mL to Liters 250. mL becomes 0.250 liters
Take the number of moles over the number of liters
0.0341 moles / 0.250 liters = 0.136 molar or 0.136M = molarity of the solution
8 0
3 years ago
For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia
Fittoniya [83]

Answer : The value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction:  Cu^{2+}+2e^-\rightarrow Cu

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

To calculate the E^o_{(Cu^{2+}/Cu)} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}

Putting values in above equation, we get:

1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)

E^o_{(Cu^{2+}/Cu)}=0.34V

Hence, the value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

8 0
3 years ago
Consider the reaction 2CuCl2 + 4K - 2Cul + 4KCI + 12. If 4 moles of CuCl2 react with 4 moles of KI, what is the limiting reactan
Free_Kalibri [48]
Haha i’m trying to do the same one i’ll make sure if i find out how too to get back to you!
8 0
3 years ago
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