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2 years ago

How many moles of lead (ii) chromate are in 51 grams of this substance? answer in units of mol?

Chemistry

1 answer:

maria [59]2 years ago

5 0

Mass of lead (II) chromate is 51 g. The molecular formula is $PbCrO_{4}$ and its molar mass is 323.2 g/mol

Number of moles can be calculated using the following formula:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Putting the values,

$n=\frac{(51 g}{323.1937 g/mol}=0.1578 mol$

Therefore, number of moles of lead (II) chromate will be 0.1578 mol.

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HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.

Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

concentration of HA, $C_A$ = 6.0mol/dm³

volume of HA, $V_A$ = 25.0cm³, = 0.025dm³

Concentration of HB, $C_B$ = 3.0mol/dm³

volume of HB, $V_B$ = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

$C_iVi = C_fV_f$

where;

$C_i$ is initial concentration

$V_i$ is initial volume

$C_f$ is final concentration of the solution

$V_f$ is final volume of the solution

$C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3$

Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

7 0

2 years ago

in a candy factory, the nutty chocolate bars contain 22.0% by mass pecans. If 5.0 kg of pecans were used for candy last Tuesday,

zysi [14]

22.0 is the same as saying that in 100 grams of a chocolate bar, there are 22.0 grams of pecans. or to make it easier because of this problem- 100 Kilograms of a chocolate bar, there is 22.0 Kg of pecans. we can use this as a conversion factor (what is used to convert a value to another value.

conversion factor---> 22.0 kg of pecan= 100 kg of chocolate bar
Note: remember this, what you are converting from goes in the denominator, what you converting to goes in the numerator.

5.0 Kg of pecan (100 Kg of chocolate bar/ 22.0 Kg of pecan)= 23 Kg of chocolate bar

7 0

2 years ago

Consider the following balanced thermochemical equation for a reaction sometimes used for H2S production:

fgiga [73]

Answer:

d. Heat is released from the reaction

Explanation:

A negative enthalpy change indicates that it is an exothermic reaction. Exothermic reactions release heat.

5 0

2 years ago

A 0.530 M Ca(OH)2 solution was prepared by dissolving 36.0 grams of Ca(OH)2 in enough water. What is the total volume of the sol

Paladinen [302]

The concentration of a solution is the number of moles of solute per fixed volume of solution.

Concentration (C) = number of moles of solute (n) / volume of the solution (v)

we have to find the volume of the solution when 36.0 g of Ca(OH)₂ is added to water to make a solution of concentration 0.530 M

mass of Ca(OH)₂ added - 36.0 g

number of moles of Ca(OH)₂ - 36.0 g / 74.1 g/mol = 0.486 mol

we know the concentration of the solution prepared and the number of moles of Ca(OH)₂ added, substituting these values in the above equation, we can find the volume of the solution

C = n/v

0.530 mol/L = 0.486 mol / V

V = 0.917 L

answer is 0.917 L

6 0

2 years ago

Read 2 more answers

How many grams of nano3 would you add to 500g of h2o in order to prepare a solution that is 0.500 molal in nano3?

VARVARA [1.3K]

When the concentration is expressed in molality, it is expressed in moles of solute per kilogram of solvent. Since we are given the mass of the solvent, which is water, we can compute for the moles of solute NaNO3.

0.5 m = x mol NaNO3/0.5 kg water
x = 0.25 mol NaNO3

Since the molar mass of NaNO3 is 85 g/mol, the mass is

0.25 mol * 85 g/mol = 21.25 grams NaNO3 needed

4 0

2 years ago