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blagie [28]
3 years ago
13

What is the molar mass of Na2CO3? 60.0 g/mol 106.0 g/mol 118.0 g/mol 141.0 g/mol

Chemistry
1 answer:
Dmitry [639]3 years ago
8 0

Answer is: the molar mass od sodium carbonate (Na₂CO₃) is 106.0 g/mol.

M(Na₂CO₃) = 2 · Ar(Na) + Ar(C) + 3 · Ar(O).

M(Na₂CO₃) = 2 · 23 + 12 + 3 · 16 · g/mol.

M(Na₂CO₃) = 46 + 12 + 48 · g/mol.

M(Na₂CO₃) = 106 g/mol; molar mass of sodium carbonate.

Ar is relative atomic mass (the ratio of the average mass of atoms of a chemical element to one unified atomic mass unit) of an element.

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For the reaction below, complete the rate expression that relates the change in concentration with respect to time to the rate o
Ann [662]

Answer: Rate in terms of disappearance of NO = -\frac{1d[NO]}{2dt}

Rate in terms of disappearance of Cl_2= -\frac{1d[Cl_2]}{1dt}

Rate in terms of appearance of NOCl = \frac{1d[NOCl]}{2dt}

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

2NO+Cl_2\rightarrow 2NOCl

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of  = -\frac{1d[NO]}{2dt}

Rate in terms of disappearance of = -\frac{1d[Cl_2]}{1dt}

Rate in terms of appearance of NOCl = +\frac{1d[NOCl]}{2dt}

5 0
3 years ago
How many grams of lithium hypochlorite (LiClO) are there in 0.594 moles?
Flauer [41]

Answer : The mass of lithium hypochlorite are, 34.7 grams.

Explanation : Given,

Moles of LiClO = 0.594 g

Molar mass of LiClO = 58.4 g/mol

Expression used :

\text{ Mass of }LiClO=\text{ Moles of }LiClO\times \text{ Molar mass of }LiClO

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

\text{ Mass of }LiClO=(0.594moles)\times (58.4g/mole)

\text{ Mass of }LiClO=34.7g

Therefore, the mass of lithium hypochlorite are, 34.7 grams.

3 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
Consider the reaction: P(s) + 5/2 Cl2(g)PCl5(g) Write the equilibrium constant for this reaction in terms of the equilibrium con
Pani-rosa [81]

Answer: The equilibrium constant for the overall reaction is K_a\times K_b

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.

a) P(s)+\frac{3}{2}Cl_2(g)\rightarrow PCl_3(g)

K_a=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}

b) PCl_3(g)+Cl_2(g)\rightarrow PCl_5(g)

K_b=\frac{[PCl_5]}{[Cl_2]\times [PCl_3]}

For overall reaction on adding a and b we get c

c) P(s)+\frac{5}{2}Cl_2(g)\rightarrow PCl_5(g)

K_c=\frac{[PCl_5]}{[Cl_2]^\frac{5}{2}}

K_c=K_a\times K_b=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}\times \frac{[PCl_5]}{[Cl_2]\times [PCl_3]}

The equilibrium constant for the overall reaction is K_a\times K_b

4 0
3 years ago
A 35 L tank of oxygen is at 315 K with an internal pressure of 190 atmospheres. How many moles of gas does the tank contain?​
VikaD [51]
N = PV = (190 atm)(35 L) = 260 moles of gas RT (0.0821 L.atm/mol.K)(315 K)
5 0
3 years ago
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