What is the formula for mass number?

Two solutions are made by mixing sugar and enough water to make a 1 liter solution. One contains 10% sugar and the other 20% sug

zysi [14]

Two solutions are made by mixing sugar and enough water to make a 1 liter solution. 10% sugar has more solvent.

In chemistry, a solution is a homogeneous mixture of two or more compounds in their relative proportions that can be continually altered up to what is referred to as the limit of solubility. Although the word "solution" is frequently used to refer to the liquid state of matter, solutions of gases and solids are also possible. A solute is dissolved by a solvent (s), which is a substance (from the Latin solv, "loosen, untie, solve,") to produce a solution. The most common form of a solvent is a liquid, although other forms include solids, gases, and supercritical fluids. Depending on the temperature, a different amount of solute can dissolve in a given volume of solvent. Solvents are extensively utilized in paints, paint removers, inks, and dry cleaning.

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7 0

1 year ago

HELPPP PLS

photoshop1234 [79]

I think it could be C maybeee though

6 0

3 years ago

How many moles of N2 you have in 476 L sample at STP?

aleksandrvk [35]

You have to take the 476 L and divide it by 22.4 because with Molar Volume 1 mol = 22.4L. After you divide it your answer is 21.25.

5 0

3 years ago

determine the freezing point depression of a solution that contains 30.7 g glycerin (c3h8o3, molar mass

Blababa [14]

The freezing point depression of a solution containing 30.7 g of glycerin is calculated as -1.65°C

Equating :

It is given that,

Given mass of glycerin is = 30.7 grams (Solute)

Volume of water = 376 mL

$K_{f}$ or molar -freezing-depression point is = 1.86°C/m

Molar mass of glycerin = 92.09 g/mole

Now, to work out the value, the mass of water should be known. Thus, to calculate, the formula used will be:

Mass = Density X Volume

Mass = 1.0 g/mL X 376 mL

Mass = 376 g or 0.376 Kg

Using the formula of melting point depression, the equation becomes:

Δ $T_{f}$ = i × $K_{f}$ ×m

T⁰- $T_{s}$ = $i *K_{f} *\frac{mass of glycerin}{molar mass of glycerin * mass of water in kg}$

in which,

Δ $T_{f}$ = change in freezing point

Δ $T_{s}$ = freezing point of solution that has to be find

ΔT° = freezing point of water ()

Since, glycerin is a non-electrolyte, the Van't Hoff factor will be 1.

Substituting the values in the above equation:

0⁰C₋T $_{s}$ = 1 ×1.86°C/m × $\frac{30.7}{92.09g/mol * 0.376kg}$

$T_{s}$ = -1.65°C

Thus, the freezing point depression of a solution is -1.65°C

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Freezing point depression

Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all less than that of the pure solvent and is directly proportional to the molality of the solute

Is melting point elevation or depression?

Boiling point elevation is that the raising of a solvent's boiling point due to the addition of a solute. Similarly, melting point depression is the lowering of a solvent's freezing point due to the addition of a solute. In fact, because the boiling point of a solvent increases, its melting point decreases

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8 0

1 year ago

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago