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Alina [70]
2 years ago
13

Given that the density of gold and sand are 19. 3 g/cm^3 and 2. 95 g/cm^3, respectively, and that the density of the mixture is

4. 17g/cm^3, calculate the percent by mass of gold in the mixture.
Chemistry
1 answer:
SOVA2 [1]2 years ago
3 0

The percent by mass of gold in the mixture will be 34.5%.

How to calculate the percentage?

The following information can be deduced based on the information given:

Density of gold = 19.3 g/cm³.

Density of sand = 2.94 g/cm³

Density of mixture = 4.17 g/cm³

It should be noted that when we find the volume fo gold on a 1cm³ sample, then we can be able to find that mass. The volume of gold will be calculated thus:

4.17 = [(19.3x + 2.95(1 - x)]

4.17 = 19.3x + 2.95 - 2.95x

4.17 = 16.35x + 2.95

x = 0.0746cm³

The mass of gold will be:

= 0.0746 × 19.3

= 1.44

The percent by mass of gold in the mixture will be:

= 1.44/4.17 × 100

= 34.5%

In conclusion, the percent by mass of gold in the mixture will be 34.5%.

What is Mass?

Mass is a physical body's total amount of matter. It also serves as a gauge for the body's inertia, or resistance to acceleration (change in velocity) in the presence of a net force. The strength of an object's gravitational pull to other bodies is also influenced by its mass.

The kilogram is the primary mass unit in the SI (kg). Even though weight is frequently measured using a spring scale rather than a balancing scale and directly compared with known masses, mass is not the same as weight in physics. Due to the lower gravity on the Moon, an object would weigh less than it does on Earth while maintaining the same mass. Given that weight is a force.

To learn more about  mass equation visit here:

brainly.com/question/19600926?referrer=searchResults

#SPJ4

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slavikrds [6]

Answer:

V_{LiOH}=21.8mL

Explanation:

Hello,

In this case, during titration at the equivalence point, we find that the moles of the base equals the moles of the acid:

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That it terms of molarities and volumes we have:

M_{LiOH}V_{LiOH}=M_{HBr}V_{HBr}

Next, solving for the volume of lithium hydroxide we obtain:

V_{LiOH}=\frac{M_{HBr}V_{HBr}}{M_{LiOH}} =\frac{0.0759M*25.0mL}{0.0896M} \\\\V_{LiOH}=21.8mL

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Which of these elements has a smaller atomic #? 1=aluminum 2=beryllium 3=tungsten
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3 years ago
Acetylsalicylic acid (aspirin), HC9H7O4, is the most widely used pain reliever and fever reducer. Find the pH of 0.035 M aqueous
KatRina [158]

<u> </u> The pH of 0.035 M aqueous aspirin is 2.48

<u>Explanation:</u>

We are given:

Concentration of aspirin = 0.035 M

The chemical equation for the dissociation of aspirin (acetylsalicylic acid) follows:

               HC_9H_7O_4\rightleftharpoons H^++C_9H_7O_4^-

<u>Initial:</u>         0.035

<u>At eqllm:</u>    0.035-x        x         x

The expression of K_a for above equation follows:

K_a=\frac{[C_9H_7O_4^-][H^+]}{[HC_9H_7O_4]}

We are given:

K_a=3.6\times 10^{-4}

Putting values in above expression, we get:

3.6\times 10^{-4}=\frac{x\times x}{(0.035-x)}\\\\x=-0.0037,0.0033

Neglecting the value of x = -0.0037 because concentration cannot be negative

So, concentration of H^+ = x = 0.0033 M

  • To calculate the pH of the solution, we use the equation:

pH=-\log[H^+]

We are given:

H^+ = 0.0033 M

Putting values in above equation, we get:

pH=-\log(0.0033)\\\\pH=2.48

Hence, the pH of 0.035 M aqueous aspirin is 2.48

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3 years ago
How would you prepare 1.00 L of a 0.400M solution of copper(II)sulfate, CuSO4?
Ber [7]
We need to dilute 0.400 mol of copper (II) sulfate, how do we know, how many weigh of CuSO_4 we have to dilute??

It's simple.

\eta=\frac{m}{MM}

Using a periodic table we can find the molar mass of Cu,~~S~~and~~O

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m=0.400*159.6

\boxed{\boxed{m=63.84~g}}

Then we have to dilute 63.84 grams of copper (II) sulfate in 1 L of water to obtain a solution with 0.400M
3 0
3 years ago
Read 2 more answers
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