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Verizon [17]
3 years ago
9

20 ml of a 1.0 m hcl solution are added to 3.0 l of acid solution with a ph of 5.35. the ph of the mixture rises to 5.33 was the

initial 3.0 l solution a buffered solution?
Chemistry
1 answer:
SashulF [63]3 years ago
3 0

You may suppose you have a 0.1 M solution of NH3, from:

NH4Cl + NaOH > NH3 + H2O.

Then you can compute the pH from the concentration of NH3 and its pKb.

The concentration is high enough to use the simplified formula:

[OH] = sqr(Kb*conc)

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Which of the following is a true statement about erosion?
ziro4ka [17]

Answer:

I would say A, Gravity can help erosion to occur.

Explanation:

While B,C,D do help cause erosion, that isn't the only way erosion can occur.

6 0
3 years ago
Read 2 more answers
The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (
Vladimir [108]

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

     [A]₀     [B]₀    [C]₀       Initial Rate (10⁻³ M/s)

1.   0.4      0.4     0.2       160

2.  0.2      0.4      0.4       80

3.   0.6     0.1       0.2       15

4.   0.2     0.1       0.2        5

5.   0.2     0.2      0.4       20

The rate of the above reaction is given as:

Rate = k[A]^{x}[B]^{y}[C]^{z}

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}

\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}

3 = 3^{x} \\\\x =1

Order w.r.t B : Use trials 2 and 5

\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}

\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}

4 = 2^{y} \\\\y =2

Order w.r.t C : Use trials 1 and 2

\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}

1 = (0.5)^{z}

z = 1

Therefore, order w.r.t C = 1

8 0
4 years ago
Hello there :3 Please anwser this Combined gas law problem for me . By the way just anwser #4 for me that’s all . I uploaded the
stellarik [79]

Answer:

136L

Explanation:

p1v1=p2v2

114 x44.0=37.0x V2

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6 0
3 years ago
What is the density of an object with volume of 5.36 mL and mass of 58.3
choli [55]

Answer:

option d is correct

Explanation:

8 0
3 years ago
Molarity of a salt water solution of "0.47" moles of NaCl dissolved in a volume of 0.25L
asambeis [7]

Answer:

1.88 M

Explanation:

The following data were obtained from the question:

Mole of NaCl = 0.47 mole

Volume of solution = 0.25L

Molarity =?

Molarity is defined as the mole of solute per unit litre of the solution. It can represented mathematically as:

Molarity = mole /Volume

Using the above formula, the molarity of the salt water solution can be obtained as follow:

Molarity = 0.47/0.25

Molarity = 1.88 M

5 0
3 years ago
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