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gladu [14]
3 years ago
5

A student puts a glass of water in the freezer. Later, he notices ice forming on the surface of the water. Which property of wat

er best explains why ice forms on its surface?
A. Ice has more surface tension than liquid water.
B. Ice is less dense than liquid water.
C. Ice has a lower freezing point than liquid water.
D. Ice is more cohesive than liquid water.
E. Ice can dissolve more oxygen than liquid water.
Chemistry
2 answers:
Blababa [14]3 years ago
6 0

Answer: Option (C) is the correct answer.

Explanation:

When a glass of water is held in the refrigerator then kinetic energy of water molecules starts to decrease as there is less temperature in the freezer.

As a result, the water starts to convert into ice because kinetic energy of water molecules converts into potential energy.

Thus, we can conclude that out of the given options, the property ice has a lower freezing point than liquid water best explains why ice forms on its surface.

Karolina [17]3 years ago
4 0
C.ice has a lower freezing point than liquid water

hope this helps
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3 years ago
Use the following equation to answer the questions below:
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Explanation:

The equation of the reaction is given as;

Be + 2HCl → BeCl2 + H2

What is the mass of beryllium required to produce 25.0g of beryllium chloride?

1 mol of Be produces 1 mol of BeCl2

Converting to mass;

Mass = Molar mass  *  Number of moles

9.01g of Be produces 79.92g of BeCl2

xg of Be produces 25g of BeCl2

Solving for x;

x = 25 * 9.01 / 79.92

x = 2.82 g

What is the mass of hydrochloric acid required to produce 25.0g of beryllium chloride? g

Converting 25.0g of beryllium chloride to moles;

Number of moles = Mass / Molar mass

Number of moles = 25 / 79.92 = 0.3128 mol

2 mol of HCl produces 1 mol of BeCl2

x mol of HCl would produce 0.3128 mol of BeCl2

solving for x;

x = 0.3128 * 2 = 0.6256 mol

Converting to mass;

Mass = 0.6256 * 36.5 = 22.83 g

What is the mass of hydrogen gas produced when 25.0g of beryllium chloride is also produced? g

25g of BeCl2 = 0.3128 mol of BeCl2

From the equation;

1 mol of H2 is produced alongside 1 mol of BeCl2

This means;

0.3128 mol of H2 would also be produced alongside 0.3128 mol of BeCl2

Mass = Number of moles * Molar mass

Mass = 0.3128mol * 2.0159 g/mol = 0.6306 g

3 0
3 years ago
Draw the Lewis structures for CH3OH, CH2O and HCOOH. Indicate the hybrid orbital used in the sigma bonds for each of the carbon
sdas [7]

Answer:

See figure 1

Explanation:

For this question, we have to remember that in the lewis structures all atoms must have<u> 8 electrons</u>. And each atom would have a different value of <u>valence electrons</u>:

Carbon => 4

Oxygen=> 6

Hydrogen=> 1

Additionally, for the <u>hybridizations</u> we have to remember that:

Sp^3=> 4 single bonds

Sp^2=> 1 double bond

Sp^1=> 1 double bond

With this in mind, the formaldehyde and formic acid would have Sp^2 carbons and the ethanol an Sp^3 carbon.

Finally, for the oxidation state. We have to remember that <u>if we have more bonds with oxygen, we will have more oxidation</u>. Therefore, the carbon that has more oxidation is the one in the formic acid (we have several bonds with oxygen).

See figure 1

I hope it helps!

4 0
3 years ago
The following shows the precipitation reaction of barium chloride (BaCl₂) and sodium hydroxide (NaOH):
Nadya [2.5K]

Answer:

The answer to your question is:

a) BaCl2

b) 0.8208 g

c) yield = 85.3 %

d)

Explanation:

                     BaCl₂(aq) + NaOH(aq) ----> Ba(OH)₂(s) + 2NaCl(aq)

Data

a) 1 g of BaCl₂

   1 g of NaOH

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MW NaOH = 23 + 16 + 1 = 40 g

                               208 g of BaCl2 -------------  1 mol

                                    1 g of BaCl2 -------------    x

                                  x = ( 1 x 1) / 208 = 0.0048 mol of BaCl2

                                    40 g of NaOH ------------  1 mol

                                       1 g of NaOH ------------   x

                                 x = (1 x 1) / 40

                                 x = 0.025 mol of NaOH

The ratio BaCl2 to NaOH is 1:1 (in the equation)

But experimentally we have 0.0048 : 0.025, so the limiting reactant is BaCl2, because is in lower concentration.

b)

                    1 mol of BaCl2 -------------- 1 mol of Ba(OH)2

                    0.0048 mol     ---------------   x

                     x = (0.0048 x 1) / 1

                     x = 0.0048 mol of Ba(OH)2

MW Ba(OH)2 = 137 + 32 + 2 = 171 g

                     171 g of Ba(OH)2 -------------------- 1 mol

                      x                         --------------------  0.0048 mol

                     x = (0.0048 x 171) / 1

                     x = 0.8208 g

c)Data

Ba(OH)2 = 0.700 g

                   % yield = 0.700 / 0.8208 x 100

                   % yield = 85.3

d)

Sorry, i don't understand this question

       

6 0
3 years ago
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san4es73 [151]
The ee solution would be 55%
6 0
3 years ago
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