Which element is a non reactive gas? Lead (82) Argon (18) Gold (79)

Aluminum has a density of 2.7 g/cm3, how much space in cm3 would 81 grams of aluminum occupy? Show steps to answering this equat

Evgesh-ka [11]

Answer:

30 cm³

Explanation:

Step 1: Given data

Density of aluminum (ρ): 2.7 g/cm³

Mass of aluminum (m): 81 g

Volume occupied by aluminum (V): ?

Step 2: Calculate the volume occupied by aluminum

The density of aluminum is equal to its mass divided by its volume.

ρ = m/V

V = m/ρ

V = 81 g / 2.7 g/cm³

V = 30 cm³

4 0

3 years ago

An object has a volume of 2 milliliters and a mass of 10 grams, calculate the density of the object

MArishka [77]

5000 kilogram/cubic meter

5 0

3 years ago

The liquid dispensed from a burette is called ___________.

11Alexandr11 [23.1K]

The liquid that is been dispensed during titration as regards this question is Titrant.

Titration can be regarded as common laboratory method that is been carried out during quantitative chemical analysis.

This analysis helps to know the concentration of an identified analyte.

Burette can be regarded as laboratory apparatus.

It is used in the in measurements of variable amounts of liquid ,this apparatus helps in dispensation of liquid, especially when performing titration.

The specifications is been done base on their volume, or resolution.

The liquid that comes out of this apparatus is regarded as Titrant, and this is gotten during titration process, which is usually carried out during volumetric analysis.

Therefore, burrete is used in volumetric analysis.

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8 0

3 years ago

Read 2 more answers

If a chemical change has taken place, which of these is most likely true?

____ [38]

C. A completely new substance is formed.

5 0

3 years ago

Read 2 more answers

In the laboratory a student combines 26.2 mL of a 0.234 M chromium(III) acetate solution with 10.7 mL of a 0.461 M chromium(III)

Natalka [10]

Answer: The molarity of $Cr^{3+}$ ions in the solution is 0.299 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For chromium (III) acetate:

Molarity of chromium (III) acetate solution = 0.234 M

Volume of solution = 26.2 mL

Putting values in equation 1, we get:

$0.234=\frac{\text{Moles of chromium (III) acetate}\times 1000}{26.2}\\\\\text{Moles of chromium (III) acetate}=\frac{0.234\times 26.2}{1000}=0.00613mol$

1 mole of chromium (III) acetate $(Cr(CH_3COO)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of acetate $(CH_3COO^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00613)=0.00613moles$

For chromium (III) nitrate:

Molarity of chromium (III) nitrate solution = 0.461 M

Volume of solution = 10.7 mL

Putting values in equation 1, we get:

$0.461=\frac{\text{Moles of chromium (III) nitrate}\times 1000}{10.7}\\\\\text{Moles of chromium (III) nitrate}=\frac{0.461\times 10.7}{1000}=0.00493mol$

1 mole of chromium (III) nitrate $(Cr(NO_3)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of nitrate $(NO_3^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00493)=0.00493moles$

For chromium cation:

Total moles of chromium cations = [0.00613 + 0.00493] = 0.01106 moles

Total volume of solution = [26.2 + 10.7] = 36.9 mL

Putting values in equation 1, we get:

$\text{Molarity of }Cr^{3+}\text{ cations}=\frac{0.01106\times 1000}{36.9}\\\\\text{Molarity of }Cr^{3+}\text{ cations}=0.299M/tex]Hence, the molarity of [tex]Cr^{3+}$ ions in the solution is 0.299 M

5 0

3 years ago