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pickupchik [31]
2 years ago
14

How many moles of CO gas are in 34.6 L?

Chemistry
1 answer:
Troyanec [42]2 years ago
6 0

moles of CO gas : 1.545

<h3>Further explanation</h3>

Standard conditions for temperature and pressure are used as a reference in certain calculations or conditions

There are 2 conditions that are usually used as a reference : STP and RTP

Assuming the STP state :

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

Then for 34.6 L of CO gas :

\tt moles=\dfrac{34.6}{22.4}=1.545

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The total concentration of ions in a 0.75 M solution of HCl is
EastWind [94]
<h3>Answer:</h3>

The total concentration of ions in a 0.75 M solution of HCl is 1.5 M

That is; 0.75 M H⁺ and 0.75 M Cl⁻

<h3>Explanation:</h3>
  • Concentration or molarity is the number of moles of a compound or an ion contained in one liter of solution. It is measured in moles per liter (M).
  • The concentration of ions making a compound is determined by the ratio of moles of the compound and the constituents ions.
  • For instance, HCl dissociates to give H⁺ and Cl⁻

       HCl(aq) → H⁺(aq) + Cl⁻(aq)

  • Therefore, since the mole ratio between HCl and the constituent ions H⁺ and Cl⁻ is 1:1, then 0.75 M of HCl dissociates to give 0.75 M H⁺ and 0.75 m Cl⁻
  • Hence the total concentration of ions in a 0.75 M solution of HCl is 1.5 M (0.75 M H⁺ and 0.75 M Cl⁻)
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A sample of nitrogen gas had a volume of 500. ml, a pressure in its closed container of 740 torr and a temperature 25 degrees c
Semenov [28]
Question:
              <span>A sample of nitrogen gas had a volume of 500mL, a pressure in its closed container of 740 torr and a temperature of 25°c. what was the volume of gas when the temperature was changed to 50°c and the new pressure was 760 torr?

Answer:

Data Given:
                   V</span>₁  =  500 mL

                   P₁  =  740 torr

                   T₁  =  25 °C + 273  =  298 K

                   V₂  =  ?

                   P₂  =  760 torr

                   T₂  =  50 °C + 273  =  323 K

Solution:
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                      P₁ V₁ / T₁  =  P₂ V₂ / T₂

Solving for V₂,

                      V₂  =  (P₁ V₁ T₂) ÷ (T₁ P₂)

Putting Values,
  
                      V₂  =  (740 torr × 500 mL × 323 K) ÷ (298 K × 760 torr)

                      V₂  =  527.68 mL
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