What was one of the first series of event in the study of chemistry

5. A 25 kg object travels a distance of 90 m in 30 s. Find the momentum

xz_007 [3.2K]

Answer:

★ v = d/t

★ v = 90/30

★ v =3

Acc to question

★ momentum = mass * velocity

★ m = 25×3

★ m = 75kgm/s

Hope it help

6 0

2 years ago

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Suppose you are presented with a clear solution of sodium thiosulfate, Na2S2O3 . How could you determine whether the solution is

navik [9.2K]

Explanation:

A solution is said to saturated when it cannot dissolve any extra solute in it. The extra solute put remains undissolved.

A solution is said to unsaturated, when the concentration of solute is less as compared to solubility of the solution it is said to be unsaturated.

A solution is said to be super saturated when it contains more of the solute than the solvent can dissolve under normal conditions is called super saturated.

5 0

3 years ago

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

Which of these pair of elements has the GREATEST electronegativity:

Georgia [21]

I think it might be Selenium.

3 0

2 years ago

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Calculate the percentage yield for the reaction represented by the equation CH4 + 2O2 ? 2H2O + CO2 when 1000 g of CH2 react with

makvit [3.9K]

Answer:

83.64%.

Explanation:

∵ The percent yield = (actual yield/theoretical yield)*100.

actual yield of CO₂ = 2300 g.

We need to find the theoretical yield of CO₂:

For the reaction:

CH₄ + 2O₂ → 2H₂O + CO₂,

1.0 mol of CH₄ react with 2 mol of O₂ to produce 2 mol of H₂O and 1.0 mol of CO₂.

Firstly, we need to calculate the no. of moles of 1000 g of CH₄ using the relation:

no. of moles of CH₄ = mass/molar mass = (1000 g)/(16.0 g/mol) = 62.5 mol.

Using cross-multiplication:

1.0 mol of CH₄ produces → 1.0 mol of CO₂, from stichiometry.

∴ 62.5 mol of CH₄ produces → 62.5 mol of CO₂.

We can calculate the theoretical yield of carbon dioxide gas using the relation:

∴ The theoretical yield of CO₂ gas = n*molar mass = (62.5 mol)(44.0 g/mol) = 2750 g.

∵ The percent yield = (actual yield/theoretical yield)*100.

actual yield = 2300 g, theoretical yield = 2750 g.

∴ the percent yield = (2300 g/2750 g)*100 = 83.64%.

5 0

3 years ago