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3 years ago

What is the salt produced when calcium hydroxide, Ca(OH)2, reacts with phosphoric acid, H3PO4, in a neutralization reaction?

Chemistry

2 answers:

ryzh [129]3 years ago

6 0

Explanation:

A neutralization reaction is a reaction in which two different compounds combine together and result in the formation of salt and water.

Foe example, $Ca(OH)_{2} + H_{3}PO_{4} \rightarrow Ca_{3}(PO_{4})_{2} + 6H_{2}O$

Therefore, salt of calcium phosphate $Ca_{3}(PO_{4})_{2}$ is formed along with formation of water when compounds $Ca(OH)_{2}$ and $H_{3}PO_{4}$ react together.

Pavel [41]3 years ago

5 0

$3Ca(OH) _{2} +2H _{3} PO _{4}$ ⇒ $6H _{2} O+Ca _{3}(PO _{4}) _{2}$
Calcium phosphate. Which is your Ca3(PO4)2.

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What is the MOLAR heat of combustion of methane(CH₄) if 64.00g of methane are burned to heat 75.0 ml of water from 25.00°C to 95

melamori03 [73]

Answer:

-5.51 kJ/mol

Explanation:

Step 1: Calculate the heat required to heat the water.

We use the following expression.

$Q = c \times m \times \Delta T$

where,

c: specific heat capacity
m: mass
ΔT: change in the temperature

The average density of water is 1 g/mL, so 75.0 mL ≅ 75.0 g.

$Q = 4.184J/g.\°C \times 75.0g \times (95.00\°C - 25.00\°C) = 2.20 \times 10^{3} J = 2.20 kJ$

Step 2: Calculate the heat released by the methane

According to the law of conservation of energy, the sum of the heat released by the combustion of methane (Qc) and the heat absorbed by the water (Qw) is zero

Qc + Qw = 0

Qc = -Qw = -22.0 kJ

Step 3: Calculate the molar heat of combustion of methane.

The molar mass of methane is 16.04 g/mol. We use this data to find the molar heat of combustion of methane, considering that 22.0 kJ are released by the combustion of 64.00 g of methane.

$\frac{-22.0kJ}{64.00g} \times \frac{16.04g}{mol} = -5.51 kJ/mol$

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3 years ago

Be sure to answer all parts. A concentration cell consists of two Sn/Sn2+ half-cells. The electrolyte in compartment A is 0.23 M

kolbaska11 [484]

Answer:

Part A. The half-cell B is the cathode and the half-cell A is the anode

Part B. 0.017V

Explanation:

Part A

The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).

So, the half-cell B is the cathode and the half-cell A is the anode.

Part B

By the Nersnt equation:

E°cell = E° - (0.0592/n)*log[anode]/[cathode]

Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.

E°cell = 0 - (0.0592/2)*log(0.23/0.87)

E°cell = 0.017V

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