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ololo11 [35]
3 years ago
8

out of 2.00 50 mL of water is added to 30 mL of a 0.50 M salt solution. What is the new concentration? (To write your answer usi

ng scientific notation use 1.0E-1 instead of 1.0 x 10-1)
Chemistry
1 answer:
vodka [1.7K]3 years ago
3 0

Answer:

The concentration of the new solution will be 0.31 M

Explanation:

The number of moles per liter of the solution is 0.50 mol. Then, in 50 ml there will be:

50 ml · (0.50 mol / 1000 ml) = 0.025 mol.

If we add 30 ml and assuming that the solution is an ideal solution, the final volume will be 80 ml.

Then, 0.025 mol will be present in 80 ml solution. In 1 l there will be:

1000 ml · (0.025 mol / 80 ml) = 0.31 mol.

The concentration of the new solution will be 0.31 M.

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We have air (21% O2 and 79% N2) at 23 bar and 30 C. 4. What is the ideal molar volume (m^3/kmol)? a. b. What is the Z factor? Wh
k0ka [10]

Answer:

The  ideal molar volume is  \frac{V}{n}  =V_z=  0.001095 \ m^3/mol  

The  Z factor is  Z = 0.09997

The  real molar volume is \frac{V_r}{n} = V_k=   0.0001095\ \frac{m^3}{mol}

Explanation:

From the question we are told that

    The pressure is  P  = 23 \ bar =  23 *10^5 Pa

      The temperature is  T  =  30 ^ oC  = 303 \ K

According to the ideal gas equation we have that

          PV  =  nRT

=>      \frac{V}{n}=V_z= \frac{RT}{P}

Where  \frac{V}{n } is the molar volume  and  R is the gas constant with value

            R  =  8.314 \ m^3 \cdot Pa \cdot K^{-1}\cdot mol^{-1}

substituting values

            \frac{V}{n}  =V_z=  \frac{ 8.314 *  303}{23 *10^{5}}

             \frac{V}{n}  =V_z=  0.001095 \ m^3/mol            

The  compressibility factor of the gas is mathematically represented  as

            Z = \frac{P *  V_z}{RT}

substituting values        

          Z = \frac{23 *10^{5} *   0.001095}{8.314 * 303}

          Z = 0.09997

Now the real molar volume is evaluated as

         \frac{V_r}{n} = V_k=  \frac{Z *  RT }{P}

substituting values

             \frac{V_r}{n} = V_k=   \frac{0.09997 *  8.314 *  303}{23 *10^{5}}

             \frac{V_r}{n} = V_k=   0.0001095\ \frac{m^3}{mol}

8 0
3 years ago
The titration of a 20.0-mLmL sample of an H2SO4H2SO4 solution of unknown concentration requires 22.87 mLmL of a 0.158 M KOHM KOH
kkurt [141]

Answer:

0.0905 M

Explanation:

Let's consider the neutralization reaction between H2SO4 and KOH.

H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O

22.87 mL of 0.158 M KOH react. The reacting moles of KOH are:

0.02287 L × 0.158 mol/L = 3.61 × 10⁻³ mol

The molar ratio of H₂SO₄ to KOH is 1:2. The reacting moles of H₂SO₄ are 1/2 × 3.61 × 10⁻³ mol = 1.81 × 10⁻³ mol

1.81 × 10⁻³ moles of H₂SO₄ are in 20.0 mL. The molarity of H₂SO₄ is:

M = 1.81 × 10⁻³ mol / 0.0200 L = 0.0905 M

3 0
3 years ago
What exactly is a proton? Is it a member of the periodic table?
Pavel [41]

The atomic number is the number of protons in the nucleus of an atom. The number of protons define the identity of an element (i.e., an element with 6 protons is a carbon atom, no matter how many neutrons may be present).

4 0
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kodGreya [7K]

ans is: 1s2, 2s2 ,2p1,...

7 0
3 years ago
2. What human-related environmental
WITCHER [35]

Answer:

Polution

Explanation:

7 0
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