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Hatshy [7]
3 years ago
15

CaCO3(s) + HCl → CaCl2(s) + CO2(g) + H2O(g)

Chemistry
1 answer:
Hoochie [10]3 years ago
4 0
HELP PLEASE I have the same question but have no clue what the answer is
You might be interested in
What is the molality of a solution that contains 1.34 moles of NaCl in 2.47 kg of solvent
dsp73

The molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.

<u>Explanation:</u>

Molality is the measure of how much of amount of solute is dissolved in the solvent. So it is calculated as the ratio of moles of solute to the grams of solvent.

         \text {Molality}=\frac{\text {Moles of solute}}{\text {Mass of solvent}}

As in this case, the solute is NaCl and solvent is unknown. So the moles of solute is given as 1.34 moles and the mass of solvent is given as 2.47 kg.

Hence, \text { molality }=\frac{1.34}{2.47}= 0.54 \mathrm{M}

Thus, the molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.

8 0
3 years ago
Copper oxide, CuO, reacts with hydrochloric acid, HCI, to produce copper chloride, CuCL2 and water
spayn [35]

Explanation:

El óxido de cobre (II), también llamado antiguamente óxido cúprico ({\displaystyle {\ce {CuO}}}{\displaystyle {\ce {CuO}}}), es el óxido de cobre con mayor número de oxidación. Como mineral se conoce como tenorita.

{\displaystyle {\ce {2Cu + O2 = 2CuO}}}{\displaystyle {\ce {2Cu + O2 = 2CuO}}}

Aquí, se forma junto con algo de óxido de cobre (I) como un producto lateral, por lo que es mejor prepararlo por calentamiento de nitrato de cobre (II), hidróxido de cobre (II) o carbonato de cobre (II):

{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}

{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}

{\displaystyle {\ce {CuCO3 = CuO + CO2}}}{\displaystyle {\ce {CuCO3 = CuO + CO2}}}

El óxido de cobre (II) es un óxido básico, así se disuelve en ácidos minerales tales como el ácido clorhídrico, el ácido sulfúrico o el ácido nítrico para dar las correspondientes sales de cobre (II):

{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}

{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}

{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}

Reacciona con álcali concentrado para formar las correspondientes sales cuprato.

{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}

Puede reducirse a cobre metálico usando hidrógeno o monóxido de carbono:

{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}

{\displaystyle {\ce {CuO + CO = Cu + CO2}}}{\displaystyle {\ce {CuO + CO = Cu + CO2}}}

6 0
3 years ago
Which of the following are excited in ultraviolet photoelectron spectroscopy?
zavuch27 [327]

Answer:

D. Valance Electrons

Explanation:

6 0
2 years ago
Consider the reaction: 2 SO2(g) + O2(g) &lt;----&gt; 2 SO3. If, at equilibrium at a certain temperature, [SO2] = 1.50 M, [O2] =
vlada-n [284]

Answer:

5.79

Explanation:

The following data were obtained from the question:

Concentration of SO2, [SO2] = 1.50 M

Concentration of O2, [O2] = 0.120 M

Concentration of SO3, [SO3] = 1.25 M

Equilibrium constant, (K) =..?

The balanced equation for the reaction is given below:

2SO2(g) + O2(g) <==> 2SO3(g)

From the balanced equation, the equilibrium constant K is written as follow:

K = [SO3]²/ [SO2]²• [O2]

With the above, the value of the equilibrium constant, K can be obtained as follow

K = [SO3]²/ [SO2]²• [O2]

K = (1.25)² / (1.50)² × 0.120

K = 1.5625 / ( 2.25 x 0.120)

K = 5.79

Therefore, the equilibrium constant is 5.79.

7 0
3 years ago
Calculate the molecular weight when a gas at 25.0 ∘C and 752 mmHg has a density of 1.053 g/L . Express your answer using three s
stiks02 [169]

Answer:

26.0 g/mol is the molar mass of the gas

Explanation:

We have to combine density data with the Ideal Gases Law equation to solve this:

P . V = n . R .T

Let's convert the pressure mmHg to atm by a rule of three:

760 mmHg ____ 1 atm

752 mmHg ____ (752 . 1)/760 =  0.989 atm

In density we know that 1 L, occupies 1.053 grams of gas, but we don't know the moles.

Moles = Mass / molar mass.

We can replace density data as this in the equation:

0.989 atm . 1L = (1.053 g / x ) . 0.082 L.atm/mol.K . 298K

(0.989 atm . 1L) / (0.082 L.atm/mol.K . 298K) = 1.053 g / x

0.0405 mol = 1.053 g / x

x =  1.053 g / 0.0405  mol = 26 g/mol

7 0
3 years ago
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