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Yuliya22 [10]
2 years ago
13

Calculate the density of a ball if you know its diameter is 15cm and its weight is approximately 2300 g. PLS HELP ILL MARK BRAIN

LIEST
Chemistry
1 answer:
dolphi86 [110]2 years ago
6 0

Answer:

575g/cm³

Explanation:

volume = 4/3 πr³

density = 2300/4/3

density = 575 g/cm³

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Reactants are<br> What you start with<br> What you end with
Naya [18.7K]

Answer:

Reactants are what you start with

Explanation:

Reactants are what you start with, the product is what you end with! :)

3 0
3 years ago
Determine whether you can swim in 1.00 x 10^27 molecules of water.​
zloy xaker [14]

Answer:

We can not swim in 1.00 × 10²⁷ molecules of water

Explanation:

The given number of molecules of water = 1.00 × 10²⁷ molecules

The Avogadro's number, N_A, gives the number of molecules in one mole of a substance

N_A ≈ 6.0221409 × 10²³ molecules/mol

Therefore

Therefore, we have;

The number of moles of water present in 1.00 × 10²⁷ molecules, n = (The number of molecules of water) ÷ N_A

∴ n = (1.00 × 10²⁷ molecules)/(6.0221409 × 10²³ molecules/mol) = 1,660.53902857 moles

The mass of one mole of water = The molar mass of water = 18.01528 g/mol

The mass, 'm', of water in 1,660.53902857 moles of water is given as follows;

Mass = (The number of moles of the substance) × (The molar mass of the substance)

∴ The mass of the water in the given quantity of water, m = 1,660.53902857 moles × 18.01528 g/mol ≈ 29.9150756 kg.

The density pf water, ρ = 997 kg/m³

Volume = Mass/Density

∴ The volume of the water present in the given quantity of water, v = 29.9150756 kg/(997 kg/m³) ≈ 30.0050909 liters

The volume of the water present in 1.00 × 10²⁷ molecules of water ≈ 30.0 liters

The average volume of a human body = 62 liters

Therefore, we can not swim in the given quantity of 1.00 × 10²⁷ molecules = 30.0 liters water

7 0
3 years ago
In a certain variety of plants, red flowers appear in majority of plants and blue flowers appear only in a few plants. Which sta
Jlenok [28]

Answer:

C

Explanation:

3 0
3 years ago
A 1.26 m aqueous solution of an ionic compound with the formula mx2 has a boiling point of 101.63 ∘c. part a calculate the van't
12345 [234]

<span>Answer is: Van't Hoff factor (i) for this solution is 1.051 .
Change in boiling point from pure solvent to solution: ΔT =i · Kb · b.
Kb - </span><span>molal boiling point elevation constant</span><span> is 0.512°C/m.
b -  molality, moles of solute per kilogram of solvent.
b = 1.26 m.
ΔT = 101.63°C - 100</span>°C = 1.63°C.

i = 1.63°C ÷ (0.512°C/m · 1.26 m).

i = 1.051.

5 0
3 years ago
Read 2 more answers
A quantity of gas occupies a volume of 804 mL at a temperature of 27 °C. At what temperature will the
irga5000 [103]

Answer:

T₂ = 150 K

Explanation:

Given data:

Initial volume of gas = 804 mL

Initial temperature = 27°C (27+273=300 K)

Final temperature = ?

Final volume = 402 mL

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁/V₁  

T₂ = 402 mL × 300 K / 804 mL

T₂ = 120,600 mL.K / 804 mL

T₂ = 150 K

3 0
3 years ago
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