A solution of salt and water is 33.0% salt by mass and has a density of 1.50 g/ml. what mass of the salt in grams is in 5.00l of this solution?

Answers

Answer 1

Answer: To answer this item, we solve first for the mass of the solution by multiplying the density by the volume. That is,

m = (density)(volume)

Substituting the known values,
m = (1.50 g/mL)(5L)(1000 mL/1L)
m = 7500 grams

To determine the mass of the salt in the solution, multiply the calculated mass of the solution by the decimal equivalent of the percent salt in the solution.

m of salt = (7500 g)(0.33)
m of salt = 2475 grams

Answer: 2475 grams

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The chemical reaction for the formation of syngas is: CH4 + H2O -> CO + 3 H2 What is the rate for the formation of hydrogen, if the rate of the formation of carbon monoxide is 0.35 M/s ? g

Answers

Answer : The rate for the formation of hydrogen is, 1.05 M/s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-(1)/(a)(d[A])/(dt)$

$\text{Rate of disappearance of B}=-(1)/(b)(d[B])/(dt)$

$\text{Rate of formation of C}=+(1)/(c)(d[C])/(dt)$

$\text{Rate of formation of D}=+(1)/(d)(d[D])/(dt)$

$Rate=-(1)/(a)(d[A])/(dt)=-(1)/(b)(d[B])/(dt)=+(1)/(c)(d[C])/(dt)=+(1)/(d)(d[D])/(dt)$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$CH_4+H_2O\rightarrow CO+3H_2$

The expression for rate of reaction :

$\text{Rate of disappearance of }CH_4=-(d[CH_4])/(dt)$

$\text{Rate of disappearance of }H_2O=-(d[H_2O])/(dt)$

$\text{Rate of formation of }CO=+(d[CO])/(dt)$

$\text{Rate of formation of }H_2=+(1)/(3)(d[H_2])/(dt)$

The rate of reaction expression is:

$\text{Rate of reaction}=-(d[CH_4])/(dt)=-(d[H_2O])/(dt)=+(d[CO])/(dt)=+(1)/(3)(d[H_2])/(dt)$

As we are given that:

$+(d[CO])/(dt)=0.35M/s$

Now we to determine the rate for the formation of hydrogen.

$+(1)/(3)(d[H_2])/(dt)=+(d[CO])/(dt)$

$+(1)/(3)(d[H_2])/(dt)=0.35M/s$

$(d[H_2])/(dt)=3* 0.35M/s$

$(d[H_2])/(dt)=1.05M/s$

Thus, the rate for the formation of hydrogen is, 1.05 M/s

Please help me...
1. oak
2. catrpiller
3. Blue jay

Answers

Answer:

$\left[\begin{array}{cc}Oak\ Leaves&Decrease\nCaterpillar&Decrease\nBlue\ Jay&Decrease\end{array}\right]$

Because the fungus is killing the oak leaves, the caterpillars will have less food and die out, resulting in a decrease in blue jay populations.

Hope it helps :) and let me know if you want me to elaborate.

Fe(ii) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution, which converts fe(ii) to insoluble fe(iii): $$4fe(oh)+(aq)+4oh−(aq)+o2(g)+2h2o(l) 4fe(oh)3(s) how many grams of o2 are consumed to precipitate all of the iron in 50.0 ml of 0.0850 m fe(ii)?

Answers

0.0340 g O2

Step 1. Write the balanced chemical equation

4Fe(OH)^(+) + 4OH^(-) + O2 + 2H2O → 4Fe(OH)3

Step 2. Calculate the moles of Fe^(2+)

Moles of Fe^(2+) = 50.0 mL Fe^(2+) × [0.0850 mmol Fe^(2+)/1 mL Fe^(2+)]

= 4.250 mmol Fe^(2+)

Step 3. Calculate the moles of O2

Moles of O2 = 4.250 mmol Fe^(2+) × [1 mmol O2/4 mmol Fe^(2+)]

= 1.062 mmol O2

Step 4. Calculate the mass of O2

Mass of O2 = 1.062 mmol O2 × (32.00 mg O2/1 mmol O2) = 34.0 mg O2

= 0.0340 g O2

0.0342 grams of O2 are consumed to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution.

To solve this problem, we need to first calculate the number of moles of Fe(II) in 50.0 mL of 0.0850 M Fe(II) solution.

Moles of Fe(II) = (0.0850 mol/L) * (50.0 mL) = 0.00425 mol

According to the balanced chemical equation, 4 moles of Fe(II) react with 1 mole of O2. Therefore, the number of moles of O2 required to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution is:

Moles of O2 = (0.00425 mol Fe(II)) * (1 mol O2 / 4 mol Fe(II)) = 0.00106 mol O2

Now we can convert the moles of O2 to grams using the molar mass of O2 (32.00 g/mol):

Grams of O2 = (0.00106 mol O2) * (32.00 g/mol) = 0.0342 g O2

Therefore, 0.0342 grams of O2 are consumed to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution.

Learn more about precipitate the iron here:

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Complete combustion of 7.80 g of a hydrocarbon produced 25.1 g of CO2 and 8.55 g of H2O. What is the empirical formula for the hydrocarbon? Insert subscripts as necessary

Answers

Answer:

The empirical formula is C3H5

Explanation:

Step 1: Data given

Mass of the compound = 7.80 grams

Mass of CO2 = 25.1 grams

Molar mass of CO2 = 44.01 g/mol

Mass of H2O = 8.55 grams

Molar mass of H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Molar mass O = 16.0 g/mol

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 25.1 grams / 44.01 g/mol

Moles CO2 = 0.570 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.570 moles CO2 we have 0.570 moles C

Step 4: Calculate mass C

Mass C = 0.570 moles * 12.01 g/mol

Mass C = 6.846 grams

Step 5: Calculate moles H2O

Moles H2O = 8.55 grams / 18.02 g/mol

Moles H2O = 0.474 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.474 moles H2O we have 2*0.474 = 0.948 moles H

Step 7: Calculate mass H

Mass H = 0.948 moles * 1.01 g/mol

Mass H = 0.957 grams

Step 8: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.570 moles / 0.570 = 1

H: 0.948 moles / 0.570 = 1.66

This means for 1 mol C we have 1.66 moles H OR for 3 moles C we have 5 moles H

The empirical formula is C3H5

Final answer:

To find the empirical formula of the hydrocarbon, divide the moles of CO2 and H2O by their molar masses. Use the smallest mole ratio to determine the empirical formula.

Explanation:

To find the empirical formula of the hydrocarbon, we need to determine the mole ratios between carbon and hydrogen in the compound. First, calculate the moles of CO2 produced by dividing the mass of CO2 by its molar mass. Next, calculate the moles of H2O produced by dividing the mass of H2O by its molar mass. Finally, divide the moles of each element by the smallest number of moles to obtain the mole ratio between carbon and hydrogen. The empirical formula is CnHm, where n and m represent the mole ratios of carbon and hydrogen, respectively.