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3 years ago

A cube of aluminum (density = 2.70 g/ml) has a mass of 17.2 g. what is the edge length of the cube?

1 answer:

8 0

The formula for mass, which is mass = density x volume, can be rewritten as volume = mass/density. We can deduce the volume of the cube as the mass and density are given: 17.2g/2.7g/ml = 6.37ml. Now that the volume of the cube is known, we can deduce the edge length. Volume = length x width x height, so 6.37 = X^3. We then take the cube root of 6.37 to discover the value of x, which will be the edge length of the cube. The edge length of the cube is 1.85 cm.

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If 72^x=26, what is the value of x?

stepan [7]

X equals 2.76923076923

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3 years ago

You have 400 kg of a radioactive substance with a short half-life of 1000 years. how much will be left after these times?

Amiraneli [1.4K]

Half-life of a radioactive substance is the time required to reduce the amount of substance to half of its initial amount.

In present case, half-life is material is given as 1000 years and initial amount of material is given as 400 kg

Answer 1) Since, half-life of radio-active substance is 1000 years, therefore after 1st half life, amount of the material will be left to half the initial amount. Hence, amount of substance left after 1000 years = 400/2 = 200 kg.

Answer 2) For 2000 years, radioactive material has crossed 2 times the half life. Therefore , amount of the material will be left to 1/4 the initial amount. Hence, amount of substance left after 2000 years = 400/4 = 100 kg.

Answer 3) For 4000 years, radioactive material has crossed 4 times the half life. Therefore , amount of the material will be left to 1/16 the initial amount. Hence, amount of substance left after 4000 years = 400/16 = 25 kg.

8 0

3 years ago

A container was found in the home of the victim that contained 120 g of ethylene glycol in 550 g of liquid. How many drinks, eac

Andrei [34K]

Answer:

0.432 drinks are toxic

Explanation:

The toxic dose of ethylene glycol is 0.1 mL per kg body weight (mL/kg). In grams (Density ethylene glycol = 1.11g/mL):

1.11g/mL * (0.1mL / kg) = 0.111g/kg

If the victim weighs 85kg, its letal dose is:

85kg * (0.111g/kg) = 9.435g of ethylene glycol

Using the concentration of ethylene glycol in the liquid:

9.435g of ethylene glycol * (550g liquid / 120g ethylene glycol) = 43.2g of liquid are toxic.

The drinks are:

43.2g of liquid * (1 drink / 100 g) =

<h3>0.432 drinks are toxic</h3>

5 0

3 years ago

The building block of all matter is a) water b) oxygen c) an atom d) an element

jeka94

My best try would be d an element

4 0

3 years ago

Show All Work For Brainliest

sattari [20]

Answer:

1) 16.8 L CO2

2) 36.96 L NH3

3) 9.88 L CO2

4) 56.99 L H2O

Explanation:

How many liters of carbon dioxide gas will be produced when 75.0 g of calcium carbonate decomposes to form calcium oxide when at STP?

CaCO3 → CaO + CO2

Moles calcium carbonate = 75.0 grams / 100.09 g/mol

Moles calcium carbonate = 0.750 moles

For 1 mol CaCO3 we'll have 1 mol CaO and 1 mol CO2

For 0.750 moles CaCO3 we'll have 0.750 moles CO2

1 mol = 22.4 L

0.750 moles CO2 = 0.750 *22.4 L = 16.8 L CO2

2. Hydrogen gas reacts with 23.1 g of nitrogen gas to produce ammonia (NH3). What volume of ammonia will be produced at STP?

3H2 + N2 → 2NH3

Moles N2 = 23.1 grams / 28.0 g/mol

Moles N2 = 0.825 moles

For 3 moles H2 we need 1 mol N2 to produce 2 moles NH3

For 0.825 moles N2 we'll have 2*0.825 = 1.65 moles NH3

1 mol = 22.4 L

1.65 mol = 1.65 * 22.4 L = 36.96 L NH3

3. Iron (III) oxide reacts with carbon monoxide to form iron and carbon dioxide. How many liters of carbon dioxide will be produced from 23.5 g of iron (III) oxide when at STP?

Fe2O3 + 3CO → 2Fe + 3CO2

Moles Fe2O3 = 23.5 grams / 159.69 g/mol

Moles Fe2O3 = 0.147 moles

For 1 mol Fe2O3 we need 3 moles CO to produce 2 moles Fe and 3 moles CO2

For 0.147 moles Fe2O3 we'll have 3*0.147 = 0.441 moles CO2

1 mol = 22.4 L

0.441 moles = 22.4 * 0.441 = 9.88 L CO2

4.How many liters of water vapor would be produced in the combustion of 12.5L of ethane, C2H6 at STP?

2C2H6 + 7O2 →4CO2 + 6H2O

22.4 L = 1 mol

12.5 L = 0.848 moles C2H6

For 2 moles C2H6 we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

For 0.848 moles C2H6 we'll have 3*0.848 = 2.544 moles H2O

1 mol = 22.4 L

2.544 moles = 22.4 L * 2.544 = 56.99 L H2O

5 0

3 years ago