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3 years ago

Find the atomicity of water.

Chemistry

1 answer:

Flauer [41]3 years ago

4 0

One molecule of water contains two atoms of hydrogen and one atom of oxygen, the atomicity of water is three.

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A 30.7 g sample of Strontium nitrate, Sr(NO3)2•nH2O, is heated to a constant mass of 22.9 g. Calculate the hydration number.

Elodia [21]

Answer:

Hydration number: 4

Explanation:

1) Mass of water in the hydrated compound

Mass of water = Mass of the hydrated sample - mass of the dehydrated compound

Mass of water = 30.7 g - 22.9 g = 7.8 g

2) Number of moles of water

Number of moles = mass in grams / molar mass

molar mass of H₂O = 2×1.008 g/mol + 15.999 g*mol = 18.015 g/mol

Number of moles of H₂O = 7.9 g / 18.015 g/mol = 0.439 mol

3) Number of moles of Strontium nitrate dehydrated, Sr (NO₃)₂

The mass of strontium nitrate dehydrated is the constant mass obtained after heating = 22.9 g

Molar mass of Sr (NO₃)₂ : 211.63 g/mol (you can obtain it from a internet or calculate using the atomic masses of each element from a periodic table).

Number of moles of Sr (NO₃)₂ = 22.9 g / 211.63 g/mol = 0.108 mol

4) Ratio

0.439 mol H₂O / 0.108 mol Sr(NO₃)₂ ≈ 4 mol H₂O : 1 mol Sr (NO₃)₂

Which means that the hydration number is 4.

4 0

3 years ago

Read 2 more answers

What volume (in liters) of a solution contains 0.14 mol of KCl?

oksano4ka [1.4K]

Answer:

$\boxed {\boxed {\sf 0.078 \ L }}$

Explanation:

We are asked to find the volume of a solution given the moles of solute and molarity.

Molarity is a measure of concentration in moles per liter. It is calculated using the following formula:

$molarity= \frac{moles \ of \ solute}{liters \ of \ solution}$

We know there are 0.14 moles of potassium chloride (KCl), which is the solute. The molarity of the solution is 1.8 molar or 1.8 moles of potassium chloride per liter.

moles of solute = 0.14 mol KCl
molarity= 1.8 mol KCl/ L
liters of solution=x

Substitute these values/variables into the formula.

$1.8 \ mol \ KCl/ L = \frac { 0.14 \ mol \ KCl}{x}$

We are solving for x, so we must isolate the variable. First, cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.

$\frac {1.8 \ mol \ KCl/L}{1} = \frac{0.14 \ mol \ KCl}{x}$

$1.8 \ mol \ KCl/ L *x = 1*0.14 \ mol \ KCl$

$1.8 \ mol \ KCl/ L *x = 0.14 \ mol \ KCl$

Now x is being multiplied by 1.8 moles of potassium chloride per liter. The inverse operation of multiplication is division, so we divide both sides by 1.8 mol KCl/L.

$\frac {1.8 \ mol \ KCl/ L *x}{1.8 \ mol \ KCl/L} = \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}$

$x= \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}$

The units of moles of potassium chloride cancel.

$x= \frac{0.14 }{1.8 L}$

$x=0.07777777778 \ L$

The original measurements of moles and molarity have 2 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 7 in the ten-thousandth place tells us to round the 7 up to a 8.

$x \approx 0.078 \ L$

There are approximately 0.078 liters of solution.

5 0

3 years ago

Which statement is true about the radii of Na and Na+?

kicyunya [14]

Answer:

Na has a larger atomic radius than Na+.

Explanation:

5 0

2 years ago

What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250

Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons $\text{H}^{+}$ than hydroxide ions $\text{OH}^{-}$ . The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that $x \; \text{L}$ of the 0.307 $\text{mol} \cdot \text{dm}^{-3}$ sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, $x$ is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain $0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x$ moles of acetic acid and $0.307 \; x$ moles of acetate ions.

Let $\text{HAc}$ denotes an acetic acid molecule and $\text{Ac}^{-}$ denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

$\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}$

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of $10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}$ . There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

$\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}$

By definition:

$\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)$

The question states that

$\text{K}_{a} = 1.75 \times 10^{-5}$

such that

$10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241$