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Zolol [24]
3 years ago
15

At standard pressure, the difference between the freezing point and the boiling point of water, in Kelvin degrees, is D. 373 A.

100 B. 180 C. 273
Chemistry
1 answer:
zepelin [54]3 years ago
6 0
Your answer would be A.100 
 
                                       

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Absolute zero is
Dmitrij [34]

Answer:

The coldest temperature possible

Explanation:

        Absolute zero =      0 K

A. is wrong.      0 °C = 273 K

B. is wrong. -210 °C =   63 K

C. is wrong.    32 °F = 273 K

E. is wrong.  -196°C =    77 K

5 0
3 years ago
For the following balanced equation: 3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l) a) How many moles of HNO3 will r
s344n2d4d5 [400]

Answer:

a) <u>26.67 moles HNO3 </u>

b) <u>0.33 moles NO</u>

c) <u>0.40 moles NO is produced</u>

d)<u>.157 moles Cu</u>

e) <u>0.105 moles NO</u>

f) <u>26.4 grams HNO3</u>

g) <u>Cu is in excess</u>

h) <u>2.41 grams Cu remain</u>

i) <u>2.37 grams NO</u>

Explanation:

Step 1: Data given

Molar mass of Cu = 63.55 g/mol

Molar mass of HNO3 = 63.01 g/mol

Molar mass of Cu(NO3)2 = 187.56 g/mol

Molar mass of NO = 30.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)

a) How many moles of HNO3 will react with 10 moles of Cu?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 10 moles Cu we need 8/3 *10 = <u>26.67 moles HNO3 </u>

b) How many moles of NO will form if 0.50 moles of Cu reacts?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.50 moles Cu we'll have 2/3 *0.50 = <u>0.33 moles NO</u>

c) If 0.80 moles of H2O forms, how much NO must also form?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

If 0.80 moles H2O is produced, 0.80/2 = <u>0.40 moles NO is produced</u>

d) How many moles of Cu are in 10.0 grams of Cu?

Moles Cu = 10.0 grams / 63.55 g/mol = 0.157 moles

In 10.0 grams Cu we have <u>0.157 moles Cu</u>

e) If 10.0 g of Cu reacts, how many moles of NO will form?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll have 2/3 * 0.157 = <u>0.105 moles NO</u>

f) If 10.0 g of Cu reacts, how many grams of HNO3 are required?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll need 0.419 moles HNO3

This is 0.419 moles * 63.01 g/mol = <u>26.4 grams HNO3</u>

g) If 10.0 g of Cu and 20.0 g of HNO3 are put together in a reaction vessel, which one will be in excess?

Moles Cu = 0.157 moles

Moles HNO3 = 20.0 grams / 63.01 g/mo = 0.317 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

The limiting reactant is HNO3. It will completely be consumed (0.317 moles). <u>Cu is in excess.</u> There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

h) How many grams of the excess substance will be left over?

There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

This is 0.038 moles * 63.55 g/mol = 2.41 grams

i) How many grams of NO will form in the reaction described in part g?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.317 moles HNO3 we'll have 0.317/4 = 0.0793 moles NO

This is 0.079 mol * 30.01 g/mol =<u> 2.37 grams NO</u>

3 0
3 years ago
The eagle has landed was said when man landed on mars or on the moon
nekit [7.7K]
The moon, man has not been to mars yet
7 0
3 years ago
A compound distributes between benzene (solvent 1) and water (solvent 2) with a distribution coefficient, K = 2.7. If 1.0g of th
mote1985 [20]

Explanation:

The given data is as follows.

Solvent 1 = benzene,          Solvent 2 = water

 K_{p} = 2.7,         V_{S_{2}} = 100 mL

V_{S_{1}} = 10 mL,       weight of compound = 1 g

       Extract = 3

Therefore, calculate the fraction remaining as follows.

                  f_{n} = [1 + K_{p}(\frac{V_{S_{2}}}{V_{S_{1}}})]^{-n}

                                  = [1 + 2.7(\frac{100}{10})]^{-3}

                                  = (28)^{-3}

                                  = 4.55 \times 10^{-5}

Hence, weight of compound to be extracted = weight of compound - fraction remaining

                                  = 1 - 4.55 \times 10^{-5}

                                  = 0.00001

or,                               = 1 \times 10^{-5}

Thus, we can conclude that weight of compound that could be extracted is 1 \times 10^{-5}.

7 0
3 years ago
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Karl went to the doctor complaining of a sore throat. The doctor performed
nydimaria [60]

Answer:

A

Explanation:

bc

6 0
3 years ago
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