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vesna_86 [32]
3 years ago
11

The alkali metals react with the halogens to form ionic metal halides. what mass of potassium chloride forms when 5.11 l of chlo

rine gas at 0.943 atm and 286 k reacts with 29.0 g potassium?
Chemistry
1 answer:
ale4655 [162]3 years ago
4 0
<span>ideal gas law: PV = nRT so .....</span><span> V = PV/(RT) </span>
<span>
Initial number of moles of Cl, n = 0.943*5.11/(0.08206 × 286) mol = 0.2053 moles.
</span><span>
We know the molar mass of K (potassium) = 39.0 g/mol </span>
<span>sooo....
The Initial number of moles of K = 29.0 g/(39.0 g/mol) = 0.7436 moles</span>

<span>Find the balanced equation for the reaction : </span><span>2K + Cl2 → 2KCl </span>
<span>Mole ratio of K:Cl = 2:1 </span>

<span>So after the reaction, the amount of K needed = (0.2053 mol) × 2 = 0.4106 mol which is less than 0.7436 mol </span>
<span>
This means that K is in excess but Cl completely reacts. </span>

<span> So we know the mole ratio is  Cl:KCl = 1 : 2 
</span>
<span>Number of moles of Cl (completely) reacted = 0.2053 mol which means the n</span><span>umber of moles of KCl formed = (0.2053 mol) × 2 = 0.4106 mol </span>

<span>Molar mass of KCl = (39.0 + 35.5) g/mol = 74.5 g/mol </span>
<span>Mass of KCl formed = 0.4106 mol * 74.5 g/mol = 30.6 g</span>
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kap26 [50]

Answer:

1) 950 mL

2) 625 mmHg

3) 426 mL

Explanation:

1) This is the relationship between pressure and volume. This relationship looks like this:

P1*V1 = P2*V2

This means the first pressure times the initial volume is equal to the second pressure times the second volume. We are solving for the second volume. First, convert the mmHg to atm and the mL to L.

1 L * 1 atm = 1.053 atm * X

X = 0.95 L or 950 mL

2) This is the same concept as the last one. :) We don't have to convert the mmHg to atm since the answer wants it in mmHg.

750 mmHg * 0.25 L = 0.3 L * X

X = 625 mmHg

3) The relationship between volume and temperature is similar to the one between pressure and temperature (like the problem in your last question). Remember to convert degrees C to Kelvin and mL to L.

V1 / T1 = V2 / T2

0.4 L / 303 K = X / 323 K

X = 0.426 L pr 426 mL

These problems become much easier once you learn the relationships between the different variables (temp, pressure, volume, etc.) When you have a problem like this, I like to first determine what relationship I am dealing with and then write out what I have and what I am solving for. This helps with organizing the problem. Then just solve it like a normal algebra problem. Always remember to convert temp to Kelvin, mL to L, and pressure to atm (unless it wants it in a different unit, then just make sure all the units match).

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Why potassium by sulphite or KHSO4 donot preserve meat
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What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )
yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution :  Given,

Concentration (c) = 0.150 M

Acid dissociation constant = k_a=4.5\times 10^{-4}

The equilibrium reaction for dissociation of HNO_2 (weak acid) is,

                           HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+

initially conc.         c                       0         0

At eqm.              c(1-\alpha)                c\alpha        c\alpha

First we have to calculate the concentration of value of dissociation constant (\alpha ).

Formula used :

k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}

Now put all the given values in this formula ,we get the value of dissociation constant (\alpha ).

4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}

4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2

0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0

By solving the terms, we get

\alpha=0.0533

No we have to calculate the concentration of hydronium ion or hydrogen ion.

[H^+]=c\alpha=0.150\times 0.0533=0.007995 M

Now we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (0.007995 M)

pH=2.097\approx 2.1

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

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