Write an equation between the carbonate and the acid

Chemistry

1 answer:

allochka39001 [22]2 years ago

8 0

Answer:

In general an acid reacts with a carbonate or hydrogen-carbonate to produce a salt, carbon dioxide gas and water.

You might be interested in

b → p experiment number [a] (m) [b] (m) initial rate (m/s) 1 0.273 0.763 2.83 2 0.273 1.526 2.83 3 0.819 0.763 25.47 22) the rat

Mazyrski [523]

The rate law for this reaction is [A]².

Balanced chemical reaction used in this experiment: A + B → P

The reaction rate is the speed at which reactants are converted into products.

Comparing first and second experiment, there is no change in initial rate. The concentration of reactant B is increased by double. Initial rate does not depands on concentration of reactant B.

Comparing first and third experiment, initial rate is nine times greater, while concentration of reactant A is three times greater. Conclusion is that concentration of reactant A is squared and the rate is [A]².

More info about rate law: brainly.com/question/16981791

#SPJ4

8 0

7 months ago

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.

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