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wel
2 years ago
5

Calculate the volume, in milliliters, for the following:

Chemistry
1 answer:
fredd [130]2 years ago
6 0

Answer: The volume for 0.850 mol of NaNO_{3} from a NaNO_{3} solution is 1700 mL.

The volume of 30.0 g of LiOH from a 2.70 M LiOH solution is 464 mL.

Explanation:

Molarity is the number of moles of solute present in a liter of solution.

  • As given moles of NaNO_{3} are 0.850 mol and molarity of NaNO_{3} solution is 0.5 M. Hence, its volume is calculated as follows.

Molarity = \frac{no. of moles}{Volume (in L)}\\0.5 M = \frac{0.850 mol}{Volume}\\Volume = 1.7 L (1 L = 1000 mL)\\= 1700 mL

Therefore, the volume for 0.850 mol of NaNO_{3} from a NaNO_{3} solution is 1700 mL.

  • As given mass of LiOH are 30.0 g from a 2.70 M LiOH (molar mass = 23.95 g/mol) solution. Hence, its number of moles are calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{30.0 g}{23.95 g/mol}\\= 1.25 mol

So, volume for LiOH solution is calculated as follows.

Molarity = \frac{no. of moles}{Volume (in L)}\\2.70 M = \frac{1.25}{Volume}\\Volume = 0.464 L (1 L = 1000 mL)\\= 464 mL

Therefore, volume of 30.0 g of LiOH from a 2.70 M LiOH solution is 464 mL.

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Which statements describe Rutherford’s model of the atom? Select three options.
AnnZ [28]

Answer:

Well, I cannot see the options but if I were you I would choose the one closest to this. Rutherford's model shows that an atom is mostly empty space, with electrons orbiting a fixed, positively charged nucleus in set, predictable paths.

Explanation:

Again I cannot see the options but here is what I would guess. Hope this helped and have a great day! :-)

6 0
2 years ago
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A sample of gas consists of 4.0 moles, and exists at stp. What is the volume of the gas?
faust18 [17]

Answer:

The volume of the gas is 89.60

Explanation:

When converting from moles to volume, you would multiply by 22.4 because 1 mole equals 22.4 liters at STP.

So...

4.0 moles * 22.4 liters =

89.60 liters

6 0
2 years ago
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What will the charge oxidiation state be for boron
artcher [175]

3+

So, compounds of boron contain boron in a positive oxidation state, generally +3. The sum of oxidation numbers of all constituent atoms of a given molecule or ion is equal to zero or the charge of the ion, respectively. ... In most of the stable compounds of boron, its oxidation number is +3

7 0
3 years ago
When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2
MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is \Delta \ T_f = T_{solvent}- T_{solution}

\Delta \ T_f = 2.9 ^0 \ C

Using the depression in freezing point, the molar depression constant of the solvent K_f = \dfrac{\Delta T_f}{m}

K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}

K_f = 4.82 ^0 C / m}

The freezing point of the solution \Delta T_f = T_{solvent} - T_{solution}

\Delta T_f = 7.3^ 0 \ C

The molality of the solution is:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

Molar depression constant of solvent X, K_f = 4.82 ^0 \ C/m

Hence, using the elevation in boiling point;

the Vant'Hoff factor i = \dfrac{\Delta T_f}{k_f \times m}

i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}

\mathbf {i = 1.51 }

3 0
2 years ago
At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---> +H2(g) + Br2 (g). If the initial partial pressures o
Damm [24]

Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of HBr = 1.0\times 10^{-2}atm

The partial pressure of H_2 = 2.0\times 10^{-4}atm

The partial pressure of Br_2 = 2.0\times 10^{-4}atm

K_p=4.2\times 10^{-9}

The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

4 0
3 years ago
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