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MArishka [77]
2 years ago
12

What is the molarity of a solution formed when you add 200. Ml of water to 50. Ml of 5. 0 m hcl

Chemistry
1 answer:
Gnesinka [82]2 years ago
6 0

The molarity of a solution formed when you add 200 mL of water to 50 mL of 5. 0 M HCl is 1.25M.

<h3>How to calculate molarity?</h3>

The molarity of a solution can be calculated using the following formula:

M1V1 = M2V2

Where;

  • M1 = molarity of a base
  • M2 = molarity of acid
  • V1 = volume of base
  • V2 = volume of acid

M1 × 200 = 50 × 5

200M1 = 250

M1 = 250/200

M1 = 1.25M

Therefore, the molarity of a solution formed when you add 200 mL of water to 50 mL of 5. 0 M HCl is 1.25M.

Learn more about molarity at: brainly.com/question/12127540

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When 1.045 g of CaO is added to 50.0 mL of water at 25.0 °C in a calorimeter, the temperature of the water increases to 32.3 °C.
Rus_ich [418]

Answer:

1.71 kJ/mol

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

\Delta H=m\times C\times \Delta T

Where,  

\Delta H  is the enthalpy change

m is the mass

C is the specific heat capacity

\Delta T  is the temperature change

Thus, given that:-

Mass of CaO = 1.045 g

Specific heat = 4.18 J/g°C

\Delta T=32.3-25.0\ ^0C=7.3\ ^0C

So,  

\Delta H=1.045\times 4.18\times 7.3\ J=31.88713\ J

Also, 1 J = 0.001 kJ

So,  

\Delta H=0.03189\ kJ

Also, Molar mass of CaO = 56.0774 g/mol

Moles=\frac{Mass}{Molar\ mass}=\frac{1.045}{56.0774}\ mol=0.01863\ mol

Thus, Enthalpy change in kJ/mol is:-

\Delta H=\frac{0.03189}{0.01863}\ kJ/mol=1.71\ kJ/mol

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Cathode-electrolyte interphase in lithium batteries revealed by cryogenic electron microscopy. matte
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It has been assumed that the intimate coating layer generated on the positive electrode, known as cathode electrolyte interphase (CEI), is crucial. However, there are still numerous questions about CEI. This results from the absence of useful instruments to evaluate the chemical and structural characteristics of these delicate interphases at the nanoscale. Here, using cryogenic electron microscopy, we establish a methodology to maintain the natural condition and directly see the interface on the positive electrode.

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Answer:

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Explanation:

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Answer:

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