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krok68 [10]
3 years ago
13

Solutions that are very concentrated have greater freezing point depression Group of answer choices true or false

Chemistry
1 answer:
Tamiku [17]3 years ago
8 0

Answer:

Explanation:

False. The greater the concentration, the lower the freezing point.

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What is the volume of a sample of mercury having a mass of 1.0 kg if the density of mercury is 13.5g/cm3?
bonufazy [111]

Answer:

<h2>volume = 74.07 cm³</h2>

Explanation:

The volume of a substance given the density and mass can be found by using the formula

volume =  \frac{mass}{Density}

From the question

Density = 13.5 g/cm³

We must first convert the mass from kg to g

1kg = 1000 g

mass of mercury = 1000 g

Substitute the values into the above formula and solve

That's

volume = \frac{1000}{13.5}

We have the final answer as

<h3>volume = 74.07 cm³</h3>

Hope this helps you

3 0
4 years ago
Which statement is true of the following reaction? H2 + O2 -&gt; 2H+ O2-
Ede4ka [16]
B.it is not balanced for charge or for number of atoms.
6 0
3 years ago
What happens to the chemical energy in methane's bonds
Alexus [3.1K]

Answer: chemical energy is converted to heat and light energy

Explanation: exothermic reaction

4CH3+7O2–>6 H2O+4CO2+energy

7 0
3 years ago
If the amount of dissolved solute in a solution at a given temperature is greater than the amount that can permanently remain in
Ksivusya [100]
I think the answer is <span>supersaturated</span>
7 0
4 years ago
Honors Stoichiometry Activity Worksheet
vaieri [72.5K]

Explanation: Here, we will be considering 1 cup is equal to 1 mol.

2 water + sugar + lemon juice → 4 lemonade

Moles of water present in 946.36 g of water=\frac{946.36 g}{236.59 g/mol}=4 mol

Moles of sugar present in 196.86 g of water=\frac{196.86 g}{225 g/mol}=0.8749 mol

Moles of lemon juice present in 193.37 g of water=\frac{193.37 g}{257.83 g/mol}=0.7499 mol

Moles of lemonade in 2050.25 g of water=\frac{2050.25 g}{719.42 g/mol}=2.8498 mol

As we can see that number of moles of lemon juice are limited.

So, we will consider the reaction will complete in accordance with moles of lemon juice.

1 mole lemon juice reacts with 2 mol of water,then 0.7499 mol of lemon juice will react with:

\frac{2}{1}\times 0.7499 mol = 1.4998 mol of water

Mass of water used =  1.4998 mol × 236.59 g/mol=354.8376 g

Water remained unused = 946.36 g - 354.8376 g =591.5223 g

1 mole lemon juice reacts with  mol of sugar,then 0.7499 mol of lemon juice will react with:

\frac{1}{1}\times 0.7499 mol = 0.7499 mol of water

Mass of sugar used =  0.7499 mol × 225 g/mol = 168.7275 g

Sugar remained unused = 196.86 g - 28.1325 g

1 mole of lemon juice gives 4 moles of lemonade.

Then 0.7499 mol of lemon juice will give:

\frac{4}{1}\times 0.7499 mol=2.996 mol of lemonade

Mass of lemonade obtained  = 2.996 mol × 719.42 g/mol = 2157.9722 g

Theoretical yield of lemonade = 2157.9722 g

Experimental yield of lemonade = 2050.25 g

Percentage yield of lemonade:

\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100

\frac{2050.25 g}{2157.9722 g}\times 100=95.00\%

The percentage yield of  lemonade is 95% and ingredient which remained unused were water and sugar.

6 0
4 years ago
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