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Zinaida [17]
3 years ago
15

Identify the chemical formula for calcium phosphate.

Chemistry
1 answer:
Aloiza [94]3 years ago
8 0

Answer:

its C for sure

Explanation:

calcium salt of phosphoric acid

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How many grams of hydrogen are contained in 2.00 mol of C6H7N
Setler [38]

Answer:- 14.0 moles of hydrogen present in 2.00 moles of [tex]C_6H_7N .

Solution:- We have been given with 2.00 moles of C_6H_7N and asked to calculate the grams of hydrogen present in it. It's a two step conversion problem. In first step we convert the moles of the compound to moles of hydrogen as one mol of the compound contains 7 moles of hydrogen. In next step the moles are converted to grams on multiplying the moles by atomic mass of H. The calculations are shown as:

2.00molC_6H_7N(\frac{7molH}{1molC_6H_7N})(\frac{1.0gH}{1molH})

= 14.0 g H

So, there are 14.0 g of hydrogen in 2.00 moles of  C_6H_7N .

3 0
3 years ago
An astronomer studying planets outside our solar system has analyzed the atmospheres of four planets. Which of these planets’ at
Nata [24]

Answer: Planet A: 76% Nitrogen, 23% Oxygen, 1% Other

Explanation: Hope this helps!

3 0
3 years ago
Which of the following will have the lowest boiling point?
mash [69]

Answer: C

Explanation:

3 0
4 years ago
Read 2 more answers
This is a type of element that has many valence electrons, not a conductor.
cestrela7 [59]
What is the question
3 0
3 years ago
A chemist fills a reaction vessel with mercurous chloride solid, mercury (I) aqueous solution, and chloride aqueous solution at
makvit [3.9K]

Answer:

ΔG° = -533.64 kJ

Explanation:

Let's consider the following reaction.

Hg₂Cl₂(s) ⇄ Hg₂²⁺(aq) + 2 Cl⁻(aq)

The standard Gibbs free energy (ΔG°) can be calculated using the following expression:

ΔG° = ∑np × ΔG°f(products) - ∑nr × ΔG°f(reactants)

where,

ni are the moles of reactants and products

ΔG°f(i) are the standard Gibbs free energies of formation of reactants and products

ΔG° = 1 mol × ΔG°f(Hg₂²⁺) + 2 mol × ΔG°f(Cl⁻) - 1 mol × ΔG°f(Hg₂Cl₂)

ΔG° = 1 mol × 148.85 kJ/mol + 2 mol × (-182.43 kJ/mol) - 1 mol × (-317.63 kJ/mol)

ΔG° = -533.64 kJ

3 0
3 years ago
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