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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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40 g of CaCO3 is how many moles of CaCO3?<br> 10 moles<br> 0.4 moles<br> 40 moles<br> 100 moles
AleksandrR [38]

Answer:

0.4 moles

Explanation:

To convert between moles and grams you need the molar mass of the compound. The molar mass of of CaCO3 is 100.09g/mol. You use that as the unit converter.

40gCaCO3* 1mol CaCO3/100.09gCaCO3 = 0.399640 mol CaCO3

This rounds to 0.4 moles CaCO3

8 0
3 years ago
What tendency is the standard reduction potential of an element based on?
Y_Kistochka [10]
I think the answer is B :)
3 0
3 years ago
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Why does mass of popcorn stay the same after popping it
Varvara68 [4.7K]

Answer:

Explanation:

it stays the same because the seeds or whatever in the bag was still the pop corn just not fully devloped

3 0
3 years ago
A gas occupies 3.5L at 2.5 kPa pressure. What is the volume at 100 mmHg at the same temperature? Be sure to
nasty-shy [4]

The volume at 100 mmHg : 0.656 L

<h3>Further explanation</h3>

Boyle's Law  

<em>At a constant temperature, the gas volume is inversely proportional to the pressure applied  </em>

\rm p_1V_1=p_2.V_2\\\\\dfrac{p_1}{p_2}=\dfrac{V_2}{V_1}

V₁=3.5 L

P₁=2.5 kPa=18,7515 mmHg

P₂=100 mmHg

\tt \dfrac{18.7515}{100}=\dfrac{V_2}{3.5}\\\\V_2=\dfrac{18.7515\times 3.5}{100}=0.656~L

4 0
2 years ago
En un matraz, disponemos de 100 g de gas oxígeno que se encuentran a 1 at de presión y 273 K de temperatura. Calcular : a) el nú
Misha Larkins [42]

Answer:

Explanation:

Dado que:

masa de oxígeno gaseoso = 100 g

presión = 1 atm

temperatura = 273 K

(a)

número de moles de oxígeno contenidos en el matraz = masa de oxígeno / masa molar de oxígeno

= 100 g / 16 gmol⁻¹

= 6.25 moles

(b) El número de moléculas de oxígeno es el siguiente:

Dado que 1 mol de oxígeno gaseoso contiene 6.023 * 10²³ moléculas de oxígeno.

Entonces, 6.25 moles contendrán:

= (6.25 ×  6.023 * 10²³) moléculas de oxígeno.

≅ 3.764 × 10²³ moléculas de oxígeno.

(c) El número de átomos de oxígeno es:

= 2 × 3.764 × 10²³

= 7.528 × 10²³ átomos de oxígeno

(d) Usando la ecuación de gas ideal

PV = nRT

El volumen ocupado por el oxígeno = \dfrac{nRT}{P}

Volumen ocupado por oxígeno = \dfrac{ 6.25 * 8.314 *273}{1}

Volumen ocupado por oxígeno= 14185.76 m³

3 0
2 years ago
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