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3 years ago

Calcium+Hydrochloric acid=

Chemistry

2 answers:

mezya [45]3 years ago

6 0

Answer:

You get Ca(II)H^2

Explanation:

This is a reaction type equation calcium reacts with HA and forms a bond.

Anastaziya [24]3 years ago

4 0

Answer:

Ca + 2HCl = CaCl2 + H2

Explanation:

Coz,

This reaction is a single displacement reaction....so, Calcium (Ca) metal is reacting with Hydrochloric acid (HCl). the Ca replaces the H in HCl and we end up with CaCl2 and H2.

Calcium + Hydrochloric acid → Calcium chloride + Hydrogen gas

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Which of the following is an example of a molecule?

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O2 is an example of a molecule

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3 years ago

Given that the molar mass of NaCl is 58.44 g/mol, Solve for the molarity of a solution that contains 87.75 g of NaCl in 500 mL o

aliina [53]

Molarity is defined as the number of moles of solute per 1 L of solvent.

the mass of NaCl in the solution is 87.75 g

number of moles of NaCl is calculated by dividing mass present by molar mass

number of NaCl moles = 87.75 g / 58.44 g/mol = 1.502 mol

the number of NaCl moles in 500 mL is - 1.502 mol

therefore number of NaCl moles in 1000 mL is - 1.502 mol/ 500 mL x 1000 mL/L = 3.004 mol

molarity of NaCl is - 3.004 M

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3 years ago

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What is an example of the strong nuclear force?

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4 years ago

Consider the reaction: CO2(g) + CCl4(g) ⇌ 2 COCl2(g) ΔG° = 46.9 kJ Under the following conditions at 25 oC: LaTeX: P_{CO_2}P C O

Mamont248 [21]

Answer : The value of $\Delta G_{rxn}$ is, $-47.0kJ/mole$

Explanation :

The formula used for $\Delta G_{rxn}$ is:

$\Delta G_{rxn}=\Delta G^o+RT\ln K_p$ ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction

$\Delta G_^o$ = standard Gibbs free energy = 46.9 kJ

R = gas constant = 8.314 J/mole.K

T = temperature = $25^oC=273+25=298K$

$K_p$ = equilibrium constant

First we have to calculate the value of $K_p$ .

The given balanced chemical reaction is,

$CO_2(g)+CCl_4(g)\rightarrow 2COCl_2(g)$

The expression for reaction quotient will be :

$K_p=\frac{(p_{COCl_2})^2}{(p_{CO_2})\times (p_{CCl_4})}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$K_p=\frac{(0.653)^2}{(0.459)\times (0.984)}$

$K_p=0.944$

Now we have to calculate the value of $\Delta G_{rxn}$ by using relation (1).

$\Delta G_{rxn}=\Delta G^o+RT\ln K_p$

Now put all the given values in this formula, we get:

$\Delta G_{rxn}=-46.9kJ/mol+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (0.944)$

$\Delta G_{rxn}=-47.0kJ/mol$

Therefore, the value of $\Delta G_{rxn}$ is, $-47.0kJ/mole$

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3 years ago

If you were asked to convert 4.2 x 1022 atoms of aluminum to moles, which of the following should you use for the conversion?

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You would divide the number of atoms by Avogadro's number [6.02x10^23]

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3 years ago