Some of the chemicals in tobacco products cause cancer. these products are called

1. Swimming pool manufacturers recommend maintaining the

Viefleur [7K]

A 3.4 × 10⁶ L swimming pool must have a mass of 1.0 × 10⁷ mg Cl₂ to maintain a concentration of 3.0 ppm.

"ppm" of "parts per million" is a unit of concentration equivalent to milligrams of solute per liters of solution.

A pool must maintain a chlorine concentration of 3.0 ppm (3.0 mg/L). The mass of chlorine in 3.4 × 10⁶ L is:

3.0 mg Cl₂/L × 3.4 × 10⁶ L = 1.0 × 10⁷ mg Cl₂

A 3.4 × 10⁶ L swimming pool must have a mass of 1.0 × 10⁷ mg Cl₂ to maintain a concentration of 3.0 ppm.

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4 0

2 years ago

Calculate the osmotic pressure associated with 50.0 g of an enzyme of molecular weight 98 g/mol dissolved in water to give 2600

andrew-mc [135]

Answer:

π = 4,882 atm

Explanation:

To calculate the osmotic pressure (π), the Van´t Hoff equation must be used, which is:

π x V = n x R x T

Where:

• π: Osmotic pressure, which is the difference between the levels of the solution and the pure solvent through a semipermeable membrane, which allows the passage of the solvent but not the solute

• V: Volume of the solution, in liters unit

• n: Number of moles of solute

• R: Constant of ideal gases, equal to 0.08206 L.atm / mol.K

• T: Absolute temperature, in Kelvin degrees

With the data you provide you can calculate the osmotic pressure by clearing it from the equation, we would be equal to:

π = (n x R x T) / V

However, all data must first be converted to the corresponding units in order to replace the values in the equation.

Solution volume ⇒ go from mL to L: 

1000 mL of solution ____ 1 L

2600 mL of solution _____ X = 2.6 L

Calculation: 2600 mL x 1 L / 1000 mL = 2.6 L

Temperature ⇒ Go from ° C to K 

T (K) = t (° C) + 273.15 = 30.0 ° C + 273.15 = 303.15 K

Number of moles of solute ⇒ It can be calculated since we have the mass of the enzyme and its molecular mass:

98.0 g of enzyme ____ 1 mol

50.0 g of enzyme _____ X = 0.510 moles

Calculation: 50.0 g x 1 mol / 98.0 g = 0.510 moles

Now, you can replace the values in the Van´t Hoff equation and you will get the result:

π = (n x R x T) / V

π = (0.510 mol x 0.08206 L.atm / mol.K x 303.15 K) / 2.6 L = 4.882 atm

Therefore, the osmotic pressure will be 4,882 atm

3 0

2 years ago

Which element can join with other elements to form covalent bonds? beryllium (Be) carbon (C) cesium (Cs) iron (Fe)

Rainbow [258]

Answer:

Carbon (C)

Explanation:

Carbon is the only nonmetal there and covalent bonds happen between two nonmetal atoms

4 0

3 years ago

Read 2 more answers

A reaction was performed in which 3.6 g 3.6 g of benzoic acid was reacted with excess methanol to make 1.3 g 1.3 g of methyl ben

Anit [1.1K]

Answer: The percent yield of the reaction is 32.34 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For benzoic acid:

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 122.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of benzoic acid}=\frac{3.6g}{122.12g/mol}=0.0295mol$

The chemical equation for the reaction of benzoic acid and methanol is:

$\text{Benzoic acid + methanol}\rightarrow \text{methyl benzoate}$

By Stoichiometry of the reaction

1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0295 moles of benzoic acid will produce = $\frac{1}{1}\times 0.0295=0.108$ moles of methyl benzoate

Now, calculating the mass of methyl benzoate from equation 1, we get:

Molar mass of methyl benzoate = 136.15 g/mol

Moles of methyl benzoate = 0.0295 moles

Putting values in equation 1, we get:

$0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol\times 136.15g/mol)=4.02g$

To calculate the percentage yield of methyl benzoate, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of methyl benzoate = 1.3 g

Theoretical yield of methyl benzoate = 4.02 g

Putting values in above equation, we get:

$\%\text{ yield of methyl benzoate}=\frac{1.3g}{4.02g}\times 100\\\\\% \text{yield of methyl benzoate}=32.34\%$

Hence, the percent yield of the reaction is 32.34 %

6 0

3 years ago

La fórmula correcta del óxido salino de estaño es: *

BaLLatris [955]

No hao luh cub chi wuh

3 0

2 years ago