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3 years ago

If the lead concentration in water is 1 ppm, then we should be able to recover 1 mg of lead from _____ L of water.

Chemistry

1 answer:

Aliun [14]3 years ago

3 0

Answer:

1 L

Explanation:

ppm means parts per million. Generally the relationship between mass and litre is given as;

1 ppm = 1 mg/L

This means that 1 ppm is equivalent to 1 mg of a substance dissolved in 1 L of water.

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Consider the reaction between a solution of molecule A and a solid block of molecule B. In general, for a reaction to occur, the

Blababa [14]

Answer:

Decreasing the volume of solvent in the solution of molecule A

Explanation:

We know that one of the factors that affect the rate of reaction is the concentration of the reactants. The greater the concentration of reactants, the faster the rate of reaction (the greater the frequency of collision between reactants).

Hence, when we decrease the volume of solvent in the solution of molecule A, the concentration of the solution increases and consequently more particles of molecule A are available to collide with particles of molecule B resulting in a higher rate of reaction.

3 0

2 years ago

How many molecules are in 41.8 g of sulfuric acid

Anton [14]

Answer

× 10²³ molecules are in 41.8 g of sulfuric acid

Explanation

The first step is to convert 41.8 g of sulfuric acid to moles by dividing the mass of sulfuric acid by its molar mass.

Molar mass of sulfuric acid, H₂SO₄ = 98.079 g/mol

$Mole=\frac{Mass}{Molar\text{ }mass}=\frac{41.8\text{ }g}{98.079\text{ }g\text{/}mol}=0.426187053\text{ }mol$

Finally, convert the moles of sulfuric acid to molecules using Avogadro's number.

Conversion factor: 1 mole of any substance = 6.022 × 10²³ molecules.

Therefore, 0.426187053 moles of sulfuric acid is equal

$\frac{0.426187053\text{ }mol}{1\text{ }mol}\times6.022×10²³\text{ }molecules=2.57\times10^{23}\text{ }molecules$

Thus, 2.57 × 10²³ molecules are in 41.8 g of sulfuric acid.

3 0

4 months ago

A chemist dissolves 867. mg of pure barium hydroxide in enough water to make up 170. mL of solution. Calculate the pH of the sol

Ilya [14]

Answer: The pH of the solution is 11.2

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml

moles of $Ba(OH)_2$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{0.867g}{171g/mol}=0.00507mol$ (1g=1000mg)

Now put all the given values in the formula of molality, we get

$Molarity=\frac{0.00507\times 1000}{170}$

$Molarity=0.0298$

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pOH=-\log [OH^-]$

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^{-}$

According to stoichiometry,

1 mole of $Ba(OH)_2$ gives 2 mole of $OH^-$

Thus 0.0298 moles of $Ba(OH)_2$ gives = $\frac{2}{1}\times 0.0298=0.0596$ moles of $OH^-$

Putting in the values:

$pOH=-\log[0.0596]=2.82$

$pH+pOH=14$

$pH=14-2.82$

$pH=11.2$

Thus the pH of the solution is 11.2

8 0

2 years ago

Find the fugacity coefficient of a gaseous species of mole fraction 0.4 and fugacity 25 psia in a mixture at a total pressure of

Oksana_A [137]

<u>Answer:</u> The fugacity coefficient of a gaseous species is 1.25

<u>Explanation:</u>

Fugacity coefficient is defined as the ratio of fugacity and the partial pressure of the gas. It is expressed as $\bar{\phi}$

Mathematically,

$\bar{\phi}_i=\frac{\bar{f_i}}{p_i}$

Partial pressure of the gas is expressed as:

$p_i=\chi_iP$

Putting this expression is above equation, we get:

$\bar{\phi}_i=\frac{\bar{f_i}}{\chi_iP}$

where,

$\bar{\phi}_i$ = fugacity coefficient of the gas

$\bar{f_i}$ = fugacity of the gas = 25 psia

$\chi_i$ = mole fraction of the gas = 0.4

P = total pressure = 50 psia

Putting values in above equation, we get:

$\bar{\phi}_i=\frac{25}{0.4\times 50}\\\\\bar{\phi}_i=1.25$

Hence, the fugacity coefficient of a gaseous species is 1.25

8 0

2 years ago

What happens to the ions when the drain is turned off? please help :] ❤️

kkurt [141]

Answer:

In a long channel MOSFET, the width of the pinch-off region is assumed small relative to the length of the channel. Thus, neither the length nor the voltage across the inversion layer change beyond the pinch-off, resulting in a drain current independent of drain bias. Consequently, the drain current saturates.

Explanation:

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5 0

2 years ago