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2 years ago

Explain what an ecosystem would look like without scavengers and decomposers.

1 answer:

6 0

Nitrogen and phosphorous would be trapped in dead remains and nutrients wouldn't return. Lots of dead material everywhere. Life would likely not continue due to lack of nutrients

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The mass percent of solute in a solution containing 3.73 g KBr dissolved in 131 g of H2O is:

Elina [12.6K]

$\huge \underbrace \mathfrak \red{Answer}$

28%

Explanation:

mass of solute(KBr) = 3.73g

mass of solvent(H2O) = 131g

mass of solution = mass of solute + mass of solvent

= 3.73 + 131

= 134.73g

$\sf \large {mass \: percentage = \frac{mass \: of \: solute}{mass \: of \: solvent} \times 100} \\ \\ \sf mass \: percentage = \frac{3.73}{134.73} \times 100 \\ \\ \sf mass \: percentage = 0.028 \times 100 \\ \\ \sf mass \: percentage = 28\%$

7 0

2 years ago

Which of these are elimination reactions? Check all that apply.

kotykmax [81]

Answer:

C3H7OH → C3H6 + H20

Explanation:

If we look at the reactant and the product we will realize that the reactant is an alcohol while the product is an alkene. The reaction involves acid catalysed elimination of water from an alcohol.

Water is a good leaving group, hence an important synthetic route to alkenes is the acid catalysed elimination of water from alcohols. Hence the conversion represented by C3H7OH → C3H6 + H20 is an elimination reaction in which water is the leaving group.

7 0

2 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