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2 years ago

The density of Silver is 10.5 g/mL. How much mass would be present given a silver cube

Chemistry

1 answer:

seraphim [82]2 years ago

3 0

Answer:

m=dxv

= 10.5 x 965 = 10132.5 rounded off to 10100 g

Explanation:

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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

3 years ago

An aqueous magnesium chloride solution is made by dissolving 7.15 moles of MgCl2 in sufficient water so that the final volume of

irinina [24]

Answer:

2.86mol/L

Explanation:

Given parameters:

Number of moles of MgCl₂ = 7.15moles

Volume of solution = 2.50L

Unknown:

Molarity of the MgCl₂ solution = ?

Solution:

The molarity of a solution is the number of moles of solute found in a given volume.

Molarity = $\frac{number of moles }{volume}$

Insert the parameters and solve;

Molarity = $\frac{7.15}{2.5}$ = 2.86mol/L

4 0

3 years ago

If the mass of 23 gummi bears is 49.2g, what is the average mass of one gummi bear? Round answer to the nearest 0.00g (do NOT ty

Mila [183]

just divide the mass 49.2 grams by the amount of gummi bears 23 to get the answer of 2.1391 then it says to round to 2 decimals so 2.14 grams

3 0

3 years ago

All of the following are forms of acid precipitation EXCEPT ________.

Aneli [31]

Dihydrogen oxide is the right answer. Dihydrogen oxide is just 2 hydrogen and 1 oxygen which is H2O or water.

6 0

3 years ago

Three mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,

8_murik_8 [283]

Answer:

(a). 46,666.7 g/mol; 78,571.4 g/mol

(b). 86950g/mol; 46,666.7 g/mol.

(c). 86950g/mol; 43,333.33 g/mol

Explanation:

So, we are given the molar masses for the three samples as: 10,000, 30,000 and 100,000 g mol−1.

Thus, the equal number of molecule in each sample = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

The average molar mass = [ ( 10,000)^2 + (30,000)^2 + 100,000)^2] ÷ 10,000 + 30,000 + 100,000 = 78,571. 4 g/mol.

(b). The equal masses of each sample = 3/[ ( 1/ 10,000) + (1/30,000 ) + (1/100,000) ] = 20930.23 g/mol.

Average molar mass = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

(c). Equal masses of the two samples = (0.145 × 10,000) + (0.855 × 100,000)/ 0.145 + 0.855 = 86950g/mol.

The weight average molar mass = 1.7 + 10,000 + 100,000/ 1.7 + 1 = 43,333.33 g/mol.

6 0

3 years ago