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Masja [62]
1 year ago
13

28.5g of iron shot is added to graduated cylinder containing 45.50mL of water. The water level rises to 49.10mL mark, from this

information, calculate the density of iron.
Mathematics
1 answer:
tankabanditka [31]1 year ago
3 0

Using the given information, the density of iron is 7.92 g/mL

<h3>Calculating the density of iron</h3>

From the question, we are to calculate the density of iron.

The density of a substance can be calculated by using the formula,

Density = Mass / Volume

From the given information,

Mass of the iron is 28.5 g

The volume of the iron is the volume displaced

Volume displaced = 49.10 mL - 45.50 mL

Volume displaced = 3.60 mL

Thus,

The density of the iron is

Density = 28.5 g / 3.60 mL

Density = 7.92 g/mL

Hence, the density is 7.92 g/mL

Learn more on Calculating the density of a substance here: brainly.com/question/11258429

#SPJ1

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Suppose brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt.
Oliga [24]

Answer:

(a) 0.288 kg/liter

(b) 0.061408 kg/liter

Step-by-step explanation:

(a) The mass of salt entering the tank per minute, x = 0.2 kg/L × 5 L/minute = 1 kg/minute

The mass of salt exiting the tank per minute = 5 × (5 + x)/500

The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;

Δ/dt = x - 5 × (5 + x)/500

The increase, in mass, Δ, after an increase in time, dt, is therefore;

Δ = (x - 5 × (5 + x)/500)·dt

Integrating with a graphing calculator, with limits 0, 10, gives;

Δ = (99·x - 5)/10

Substituting x = 1 gives

(99 × 1 - 5)/10 = 9.4 kg

The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank)

∴ The concentration of the salt and water in the tank after 10 minutes =  (5 + 9.4)/500 = (14.4)/500 = 0.288

The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter

(b) With the added leak, we now have;

Δ/dt = x - 6 × (14.4 + x)/500

Δ = x - 6 × (14.4 + x)/500·dt

Integrating with a graphing calculator, with limits 0, 20, gives;

Δ = 19.76·x -3.456 = 16.304

Where x = 1

The increase in mass after an increase in = 16.304 kg

The total mass = 16.304 + 14.4 = 30.704 kg

The concentration of the salt in the tank then becomes;

Concentration = 30.704/500 = 0.061408 kg/liter.

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Answer:

Step-by-step explanation:

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