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3 years ago

An unknown weak acid with a concentration of 0.086 M has a pH of 1.80. What is the Ka of the weak acid?

Chemistry

1 answer:

AlekseyPX3 years ago

5 0

Answer:

2.90 × 10⁻³

Explanation:

Step 1: Given data

pH: 1.80

Concentration of the weak acid (Ca): 0.086 M

Step 2: Calculate the concentration of H⁺

We will use the following expression.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -1.80

[H⁺] = 0.0158 M

Step 3: Calculate the acid dissociation constant (Ka)

We will use the following expression.

Ka = [H⁺]² / Ca

Ka = 0.0158² / 0.086

Ka = 2.90 × 10⁻³

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HNO3 + S --> H2SO4 + NO Now identify the element oxidized and the element reduced. Which element is oxidized? Which element i

OleMash [197]

Answer: S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

Explanation:

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when oxidation number of a species decreases.

For the given chemical reaction:

$HNO_3+S\rightarrow H_2SO_4+NO$

On the reactant side:

Oxidation number of H = +1

Oxidation number of N = +5

Oxidation number of O = -2

Oxidation number of S = 0

On the product side:

Oxidation number of H = +1

Oxidation number of N = +2

Oxidation number of O = -2

Oxidation number of S = +6

As the oxidation number of S is increasing from 0 to +6. Thus, it is getting oxidized. Similarly, the oxidation number of N is decreasing from +5 to +2. Thus, it is getting reduced.

The oxidation numbers of O and H remain the same on both sides of the reaction. Thus, they are neither getting oxidized or reduced.

Hence, S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

3 0

2 years ago

What happens to a line when the y-intercept is changed? Check all that apply.

tamaranim1 [39]

Answer:

As the y-intercept increases, the graph of the line shifts up;

As the y-intercept decreases, the graph of the line shifts down

Explanation:

There are two ways to think about this problem. The first way would be the graphical approach:

if we only change the y-intercept, this means we keep the same slope;
y-axis is the vertical axis;
if we change the point at which the line crosses the y-axis, we either shift it upward for a higher y-intercept or downward for a lower y-intercept.

Now, thinking algebraically, a line has the following equation in a general form:

$y = mx + b$

The y-intercept is essentially obtained when x = 0, then:

y = b:

if we increase b value, the y value increases, so the graph shifts upward;
if we decrease b value, the y value decreases, so the graph shifts downward.

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3 years ago

Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50

devlian [24]

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C $_{V}$ You can work out C $_{V}$

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC $_{V}$ ∆T

Polyatomic gas: C $_{V}$ = 3R

∆U= nC $_{V}$ ∆T

∆U= 28g x C $_{V}$ x (350K - 58.3K)

∆U = 28C $_{V}$ x 291.7

∆U = 10967.6 x C $_{V}$

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3 years ago

Enter a range of values for x, 2x+10, 14

PIT_PIT [208]

Answer:

62 degrees and 15.

Explanation:

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3 years ago

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Paladinen [302]

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3 years ago