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4 years ago

Which of the following solutions would have the lowest freezing point? 1.0 nacl 3.0 nacl 2.0 nacl or 5.0 nacl

2 answers:

5 0

The solution that will have the lowest freezing point is 5.0 SODIUM CHLORIDE.
Adding solute to solvents usually result in the depression of the freezing point. The higher the quantity of the solute that is added, the lower the freezing point of the solution.

Elena-2011 [213]4 years ago

5 0

you typed it in wrong, it should be D.) 0.5 NaCI as a choice.

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Convert 3.8 dm3 to the unit m3.

Artist 52 [7]

Are you an flvs student?

5 0

3 years ago

Read 2 more answers

The limiting reactant in a reaction a. is the reactant for which there is the least amount in grams b. is the reactant which has

Shalnov [3]

Actually the correct answer must be:

The limiting reactant in the reaction is the one which has the lowest ratio of moles available over coefficient in the balanced equation

This is because the actual mass or number of moles of the reactant does not directly dictate if it is a limiting reactant, this must be relative to the other reactants.

So the answer is:

e. none of the above

5 0

3 years ago

Dimethyl sulfoxide [(ch3)2so], also called dmso, is an important solvent that penetrates the skin, enabling it to be used as a t

Bas_tet [7]

The molecular formula of dimethyl sulfoxide is $(CH_{3})_{2}SO$ . Molar mass of dimethyl sulfoxide is 78.13 g/mol. Calculate number of moles as follows:

$n=\frac{m}{M}=\frac{7.14\times 10^{3} g}{78.13 g/mol}=91.38 mol$

From the molecular formula, 1 mole of dimethyl sulfoxide contains 2 moles of Carbon, 6 moles of Hydrogen, 1 mole of Sulfur and 1 mole of oxygen.

Thus, 91.38 moles of dimethyl sulfoxide will have:

Carbon :

$n_{C}=2\times 91.38 moles=182.77 moles$

Hydrogen:

$n_{H}=6\times 91.38 moles=548.28 moles$

Sulfur:

$n_{S}=1\times 91.38 moles=91.38 moles$

Oxygen:

$n_{O}=1\times 91.38 moles=91.38 moles$

Since, 1 mole of an element equals to $6.023\times 10^{23}$ atoms thus, number of atoms can be calculated as:

Carbon:

$1 mole\rightarrow 6.023\times 10^{23} atoms$

Thus,

$182.77 moles\rightarrow 182.77\times 6.023\times 10^{23} atoms=1.10\times 10^{26} atoms$

Hydrogen:

$1 mole\rightarrow 6.023\times 10^{23} atoms$

Thus,

$548.28 moles\rightarrow 548.28\times 6.023\times 10^{23} atoms=3.30\times 10^{26} atoms$

Sulfur:

$1 mole\rightarrow 6.023\times 10^{23} atoms$

Thus,

$91.38 moles\rightarrow 91.38\times 6.023\times 10^{23} atoms=5.50\times 10^{25} atoms$

Oxygen:

$1 mole\rightarrow 6.023\times 10^{23} atoms$

Thus,

$91.38 moles\rightarrow 91.38\times 6.023\times 10^{23} atoms=5.50\times 10^{25} atoms$

Therefore, number of C, S, H and O atoms are $1.10\times 10^{26}, 5.50\times 10^{25}, 3.30\times 10^{26} and 5.50\times 10^{25} atoms$ respectively.

4 0

4 years ago

H2o has how many atoms in it? 1,2, or 3?

liraira [26]

H2o only has two atoms in it.
(just trying to make this text longer,,,)

4 0

3 years ago

To what volume should you dilute 122 mL of an 8.20 M CuCl2 solution so that 51.0 mL of the diluted solution contains 4.40 g CuCl

BartSMP [9]

Answer:

<h2>The first thing to do here is to use the molarity and the volume of the initial solution to figure out how many grams of copper(II) chloride it contains.</h2><h2 /><h2>133</h2><h2>mL solution</h2><h2>⋅</h2><h2>1</h2><h2>L</h2><h2>10</h2><h2>3</h2><h2>mL</h2><h2>⋅</h2><h2>7.90 moles CuCl</h2><h2>2</h2><h2>1</h2><h2>L solution</h2><h2>=</h2><h2>1.051 moles CuCl</h2><h2>2</h2><h2 /><h2>To convert this to grams, use the compound's molar mass</h2><h2 /><h2>1.051</h2><h2>moles CuCl</h2><h2>2</h2><h2>⋅</h2><h2>134.45 g</h2><h2>1</h2><h2>mole CuCl</h2><h2>2</h2><h2>=</h2><h2>141.31 g CuCl</h2><h2>2</h2><h2 /><h2>Now, you know that the diluted solution must contain </h2><h2>4.49 g</h2><h2> of copper(II) chloride. As you know, when you dilute a solution, you increase the amount of solvent while keeping the amount of solute constant.</h2><h2 /><h2>This means that you must figure out what volume of the initial solution will contain </h2><h2>4.49 g</h2><h2> of copper(II) chloride, the solute.</h2><h2 /><h2>4.49</h2><h2>g</h2><h2>⋅</h2><h2>133 mL solution</h2><h2>141.32</h2><h2>g</h2><h2>=</h2><h2>4.23 mL solution</h2><h2>−−−−−−−−−−−−−− </h2><h2 /><h2>The answer is rounded to three sig figs.</h2><h2 /><h2>You can thus say that when you dilute </h2><h2>4.23 mL</h2><h2> of </h2><h2>7.90 M</h2><h2> copper(II) chloride solution to a total volume of </h2><h2>51.5 mL</h2><h2> , you will have a solution that contains </h2><h2>4.49 g</h2><h2> of copper(II) chloride.</h2>

3 0

3 years ago