A group on the periodic table is best described as a ________

What is the concentration of a solution with a volume of 2.5 liters containing 600 grams of calcium phosphate?

trapecia [35]

Answer:

1.12M

Explanation:

Given parameters:

Volume of solution = 2.5L

Mass of Calcium phosphate = 600g

Unknown:

Concentration = ?

Solution:

Concentration is the number of moles of solute in a particular solution.

Now, we find the number of moles of the calcium phosphate from the given mass;

Formula of calcium phosphate = Ca₃PO₄

molar mass = 3(40) + 31 + 4(16) = 215g/mol

Number of moles of Ca₃PO₄ = $\frac{600}{215}$ = 2.79moles

Now;

Concentration = $\frac{Number of moles }{volume }$

Concentration = $\frac{2.79}{2.5}$ = 1.12M

7 0

3 years ago

Question 2 of 50

wolverine [178]

The thermal decomposition of calcium carbonate will produce 14 g of calcium oxide. The stoichiometric ratio of calcium carbonate to calcium oxide is 1:1, therefore the number of moles of calcium carbonate decomposed is equal to the number of moles of calcium oxide formed.

Further Explanation:

To solve this problem, follow the steps below:

Write the balanced chemical equation for the given reaction.
Convert the mass of calcium carbonate into moles.
Determine the number of moles of calcium oxide formed by using the stoichiometric ratio for calcium oxide and calcium carbonate based on the coefficient of the chemical equation.
Convert the number of moles of calcium oxide into mass.

Solving the given problem using the steps above:

STEP 1: The balanced chemical equation for the given reaction is:

$CaCO_{3} \rightarrow \ CaO \ + \ CO_{2}$

STEP 2: Convert the mass of calcium carbonate into moles using the molar mass of calcium carbonate.

$mol \ CaCO_{3} \ = 25 \ g \ CaCO_{3} \ (\frac{1 \ mol \ CaCO_{3}}{100.0869 \ g \ CaCO_{3}})\\ \\\boxed {mol \ CaCO_{3} \ = 0.2498 \ mol}$

STEP 3: Use the stoichiometric ratio to determine the number of moles of CaO formed.

For every mole of calcium carbonate decomposed, one more of a calcium oxide is formed. Therefore,

$mol \ CaO \ = 0.2498 \ mol$

STEP 4: Convert the moles of CaO into mass of CaO using its molar mass.

$mass \ CaO \ = 0.2498 \ mol \ CaO \ (\frac{56.0774 \ g \ CaO}{1 \ mol \ CaO})\\ \\mass \ CaO \ = 14.008 \ g$

Since there are only 2 significant figures in the given, the final answer must have the same number of significant figures.

Therefore,

$\boxed {mass \ CaO \ = 14 \ g}$

Learn More

Learn more about stoichiometry brainly.com/question/12979299
Learn more about mole conversion brainly.com/question/12972204
Learn more about limiting reactants brainly.com/question/12979491

Keywords: thermal decomposition, stoichiometry

5 0

3 years ago

Consider the chemical equations shown here.

vichka [17]

P₄0₆ $_{s}$ + 20₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$

Explanation:

The overall equation for the reaction that produces P₄0₁₀ is :

P₄0₆ $_{s}$ + 20₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$

Now let us derive this equation:

Given equations:

P₄ $_{s}$ + 30₂ $_{g}$ ⇒ P₄0₆ $_{s}$ equation 1;

P₄ $_{s}$ + 50₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$ equation 2;

To get the overall combined equation, the equation 1 must be reversed and added to equation 2:

P₄0₆ $_{s}$ ⇒ P₄ $_{s}$ + 30₂ $_{g}$ equation 3

+

equation 2:

P₄ $_{s}$ + 50₂ $_{g}$ + P₄0₆ $_{s}$ ⇒ P₄0₁₀ $_{s}$ + P₄ $_{s}$ + 30₂ $_{g}$

cancelling specie that appears on both sides and removing excess oxygen gas on the reactant side gives;

P₄0₆ $_{s}$ + 20₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$

learn more:

Net equation brainly.com/question/2947744

#learnwithBrainly

5 0

3 years ago

The atmosphere is composed of

allsm [11]

I think it's the c (<span>.all the gases that surround the Earth)</span>

8 0

3 years ago

How many C-13 atoms are present, on average, in a 1.6000×104-atom sample of carbon?

san4es73 [151]

I don't know why I am answering this question but assuming C-13 has a natural abundance of 1.07%:

(1.6000x10^4)(0.0107) = 171.2 = 171 atoms of C-13

5 0

3 years ago

Read 2 more answers

A group on the periodic table is best described as a _________.