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xxTIMURxx [149]
2 years ago
10

You work for a cutlery manufacturer who wants to electrolytically precipitate 0.500 g of silver onto each piece of a batch of 25

0 forks. The preferred electrolytic solution for silver is AgCN(aq). Aqueous AgCN is purchased as a 2.50 M solution. How many mL of AgCN(aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks
Chemistry
1 answer:
ICE Princess25 [194]2 years ago
6 0

Answer:

373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.

Explanation:

Mass of silver to be precipitated on ecah spoon = 0.500 g

Number of silver spoons = 250

Total mass of silver = 250 × 0.500 g = 125 g

Moles (n)=Molarity(M)\times Volume (L)

Moles of AgCN = n  = \frac{125 g}{134 g/mol}=0.9328 mol

Volume of AgCN solution =V

Molarity of the AgCN = 2.50 M

V=\frac{0.9328 mol}{2.50 M}=0.3731 L=373.1 mL

(1 L = 1000 mL)

373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.

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Which has more momentum- an 80 kg runner with a velocity of 2.45 m/s or a 65 kg runner with velocity of 3 m/s? Show work
castortr0y [4]
Momentum = Mass x Velocity
80kg Runner: 80 x 2.45 = 196 Kg m/s
65kg Runner: 65 x 3= 195 Kg m/s

The 80kg runner has a greater momentum
8 0
2 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
2 years ago
An unopened, cold 2.00 L bottle of soda contains 46.0 mL of gas confined at a pressure of 1.30 atm at a temperature of 5.0oC. If
Katena32 [7]

Answer:

20.8mL

Explanation:

7 0
2 years ago
Lithium has an atomic mass of 6.941 amu. Lithium has two common isotopes. The one isotope has a mass of 6.015 amu and a relative
koban [17]

Answer:

The atomic mass of second isotope is 7.016

Explanation:

Given data:

Average Atomic mass of lithium = 6.941 amu

Atomic mass of first isotope = 6.015 amu

Relative abundance of first isotope = 7.49%

Abundance of second isotope = ?

Atomic mass of other isotope = ?

Solution:

Total abundance = 100%

100 - 7.49 = 92.51%

percentage abundance of second isotope = 92.51%

Now we will calculate the mass if second isotope.

Average atomic mass of lithium = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

6.941 = (6.015×7.49)+(x×92.51) /100

6.941 =  45.05235 + (x92.51) / 100

6.941×100 = 45.05235 + (x92.51)

694.1 - 45.05235   = (x92.51)

649.04765 = x 92.51

x = 485.583 /92.51

x = 7.016

The atomic mass of second isotope is 7.016

3 0
2 years ago
If a 60-g object has a volume of 30 cm3, what is its density? 2 g/cm3 0.5 cm3/g 1800 g * cm3 none of the above
jasenka [17]

Answer : The density of an object is, 2g/cm^3

Solution : Given,

Mass of an object = 60 g

Volume of an object = 30cm^3

Formula used :

\text{Density of an object}=\frac{\text{Mass of an object}}{\text{Volume of an object}}

Now put all the given values in this formula, we get the density of an object.

\text{Density of an object}=\frac{60g}{30cm^3}=2g/cm^3

Therefore, the density of an object is, 2g/cm^3

6 0
1 year ago
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