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Kazeer [188]
3 years ago
5

Calculate the pH of a 0.10 M HCN solution that is 0.0070% ionized.

Chemistry
1 answer:
Anastaziya [24]3 years ago
7 0

Answer:

D) 5.15

Explanation:

Step 1: Write the equation for the dissociation of HCN

HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)

Step 2: Calculate [H⁺] at equilibrium

The percent of ionization (α%) is equal to the concentration of one ion at the equilibrium divided by the initial concentration of the acid times 100%.

α% = [H⁺]eq / [HCN]₀ × 100%

[H⁺]eq = α%/100% × [HCN]₀

[H⁺]eq = 0.0070%/100% × 0.10 M

[H⁺]eq = 7.0 × 10⁻⁶ M

Step 3: Calculate the pH

pH = -log [H⁺] = -log 7.0 × 10⁻⁶ = 5.15

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14.7 grams of magnesium reacts completely with 9.7 grams of oxygen to form magnesium oxide (MgO). What is the percent compositio
swat32
The answer is 60.3% magnesium, 39.7% oxygen.
Solution:
The chemical equation for the reaction is 2 Mg + O2 → 2 MgO.
Since magnesium reacts completely with oxygen, it is the limiting reactant in the reaction. Hence, we can use the number of moles of magnesium to get the mass of MgO produced:
     moles of magnesium = 14.7g / 24.305g mol-1
                                       = 0.6048 mol

     mass of MgO = 0.6048mol Mg(2 mol MgO/2mol Mg)(40.3044g MgO/1 mol MgO) 
                           = 24.376g MgO

We can now solve for the percentage of magnesium:
     % Mg = (14.7g Mg / 24.376g MgO)*100% = 60.3%

We also use the number of moles of magnesium to get the mass of oxygen consumed in the reaction:
     mass of O2 =  0.6048 mol Mg (1mol O2 / 2mol Mg) (31.998g / 1mol O2) 
                        = 9.676g

The percentage of oxygen is therefore
     % O2 = (9.676g O2 / 24.376g MgO)*100%
               = 39.7%
Notice that we can just subtract the magnesium's percentage from 100% to get 
     % O2 = 100% - 60.3% = 39.7%
5 0
3 years ago
A chemist observes that a large molecule reacts as if it were much smaller. The chemist proposes that the molecule is folded in
nadezda [96]

Answer:

Hypothesis

Explanation:

The following steps are applicable when we wish to prove a specific fact:

  • a hypothesis is made; this is a statement that we provide after some observations and we wish to either prove or deny it;
  • multiple experiments are carried out in order to gather significantly substantial amount of data that can be then further analyzed and any tendencies can be noticed;
  • based on the data gathered, conclusions are made: we either prove or deny the hypothesis. If hypothesis is proved, it may become a theory over long time.

In the context of this problem, we're at the first step where we make a hypothesis.

6 0
3 years ago
Jose’s lab instructor gives him a solution of sodium phosphate that is buffered to a pH of 4. Because of an error that he made w
RoseWind [281]
His measurements are precise since his pH values are close to each other in a way that it was repeated in all measurements. On the contrary to accuracy, it is the closeness to the actual pH value he should have achieved. Therefore, Jose's results are precise but not accurate since his value is not close to the actual value of pH 4.
3 0
3 years ago
onvert the value of Kc to a value of Kp for the following reaction: N2(g)+3H2(g)⇌2NH3(g) Kc = 0.50 at 400 °C.
ale4655 [162]

Answer:

K_p= 0.00016

Explanation:

The relation between Kp and Kc is given below:

K_p= K_c\times (RT)^{\Delta n}

Where,  

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

N_2_{(g)}+3H_2_{(g)}\rightleftharpoons2NH_3_{(g)}

Given: Kc = 0.50

Temperature = 400^oC=[400+273]K=673K

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (2)-(3+1) = -2

Thus, Kp is:

K_p= 0.50\times (0.082057\times 673)^{-2}

K_p= 0.00016

6 0
3 years ago
What temperature would 3.54 moles of xenon gas need to reach to exert a pressure of 1.57 atm at a volume of 34.6 l
ira [324]

Answer:

186.9Kelvin

Explanation:

The ideal gas law equation is PV = n R T

where

P   is the pressure of the gas

V   is the volume it occupies

n  is the number of moles of gas present in the sample

R  is the universal gas constant, equal to  0.0821 atm L /mol K

T  is the absolute temperature of the gas

Ensure units of the volume, pressure, and temperature of the gas correspond to R ( the universal gas constant, equal to  0.0821 atm L /mol K )

n = 3.54moles

P= 1.57

V= 34.6

T=?

PV = n R T

PV/nR = T

1.57 x 34.6/3.54 x 0.0821

54.322/0.290634= 186.908620464= T

186.9Kelvin ( approximately to 1 decimal place)

5 0
3 years ago
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