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Margaret [11]
3 years ago
14

HNO3 is an Arrhenius ______________ and increases the concentration of __________ when added to water.

Chemistry
1 answer:
poizon [28]3 years ago
4 0
HNO3 and H2SO4 are Arrhenius acids which will increase the concentration of H+ when dissolved in water.

KOH and Ca(OH)2 are Arrhenius bases that increase the concentration of OH- when dissociated in water.

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A major component of gasoline is octane, C8H18. When octane is burned in air, it chemically reacts with oxygen gas (O2) to produ
ddd [48]

Answer:

m_{CO_2}=2.8gCO_2

Explanation:

Hello,

In this case, the combustion of octane is chemically expressed by:

C_8H_{18}+\frac{25}{2} O_2\rightarrow 8CO_2+9H_2O

In such a way, due to the 25/2:8 molar ratio between oxygen and carbon dioxide, we can compute the yielded grams of carbon dioxide (molar mass 44 g/mol) as shown below:

m_{CO_2}=3.2gO_2*\frac{1molO_2}{32gO_2} *\frac{8molCO_2}{\frac{25}{2}molO_2 } *\frac{44gCO_2}{1molCO_2}\\ \\m_{CO_2}=2.8gCO_2

Best regards.

3 0
3 years ago
How do I solve this?
ira [324]
Sorry, I can't really see the question )-:
6 0
4 years ago
A 0.595 g sample of a metal, M, reacts completely with sulfuric acid according to the reaction M ( s ) + H 2 SO 4 ( aq ) ⟶ MSO 4
zalisa [80]

Answer:

molar mass M(s) = 65.326 g/mol

Explanation:

  • M(s) + H2SO4(aq) → MSO4(aq) + H2(g)

∴ VH2(g) = 231 mL = 0.231 L

∴ P atm = 1.0079 bar

∴ PvH2O(25°C) = 0.03167 bar

Graham´s law:

⇒ PH2(g) = P atm - PvH2O(25°C)

⇒ PH2(g) = 1.0079 bar - 0.03167 bar = 0.97623 bar = 0.9635 atm

∴ nH2(g) = PV/RT

⇒ nH2(g) = ((0.9635 atm)(0.231 L))/((0.082 atmL/Kmol)(298 K))

⇒ nH2(g) = 9.1082 E-3 mol

⇒ n M(s) = ( 9.1082 E-3 mol H2(g) )(mol M(s)/mol H2(g))

⇒ n M(s) = 9.1082 E-3 mol

∴ molar mass M(s) [=] g/mol

⇒ molar mass M(s) =   (0.595 g) / (9.1082 E-3 mol)

⇒ molar mass M(s) = 65.326 g/mol

7 0
3 years ago
When the following equilibrium process
ElenaW [278]

Answer:

K(eq) = 15  (2 sig. figs)

Explanation:

Rxn:         CO(g)    +    2H₂(g)    ⇄   CH₃OH(l)

C(eq):    0.150M       0.360M           0.282M  

Keq = [CH₃OH(l)]/[CO(g)][H₂(g)]

= (0.282M)/(0.150M)(0.360M)²  

= 14.50617284 (calc. ans.)

= 15 (2 sig. figs.)  

4 0
3 years ago
Account for the change when NO2Cl is added using the reaction quotient Qc. Match the words in the left column to the appropriate
ryzh [129]

Answer:

A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc.

B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases.

C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases.

Explanation:

Hello,

In this case, for the equilibrium reaction:

NO_2Cl(g)+NO(g)\rightleftharpoons NOCl(g)+NO_2(g)

Whose equilibrium expression is:

Kc=\frac{[NO_2][NOCl]}{[NO_2Cl][NO]}

The proper matching is:

A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc, since the denominator becomes greater, therefore, Qc decreases.

B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases, since the lower the denominator, the higher Qc as it has the concentration of reactants.

C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases, since the reactants must be consumed in order to reestablish equilibrium by shifting the reaction towards the products.

Best regards.

8 0
4 years ago
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