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Tems11 [23]
3 years ago
6

Calculate the mass of vanadium(V) oxide (V2O5) that contains a billion (1.000e9) vanadium atoms. Be sure your answer has a unit

symbol if necessary, and round it to 4 significant digits._______
Chemistry
1 answer:
Finger [1]3 years ago
7 0

Answer:

1.510 × 10⁻¹³ g

Explanation:

Step 1: Given data

Number of atoms of vanadium: 1.000 × 10⁹ atoms

Step 2: Calculate the molecules of V₂O₅ that contain 1.000 × 10⁹ atoms of V

Each molecule of V₂O₅ has 2 atoms of V.

1.000 × 10⁹ atom V × (1 molecule V₂O₅/2 atom V) = 0.5000 × 10⁹ molecule V₂O₅

Step 3: Calculate the moles corresponding to 0.5000 × 10⁹ molecules of V₂O₅

We will use Avogadro's number: there are 6.022 × 10²³ molecules of V₂O₅ in 1 mole of V₂O₅.

0.5000 × 10⁹ molecule × (1 mol/6.022 × 10²³ molecule) = 8.303 × 10⁻¹⁶ mol

Step 4: Calculate the mass corresponding to 8.303 × 10⁻¹⁶ moles of V₂O₅

The molar mass of V₂O₅ is 181.88 g/mol.

8.303 × 10⁻¹⁶ mol × 181.88 g/mol = 1.510 × 10⁻¹³ g

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Enough of a monoprotic weak acid is dissolved in water to produce a 0.0144 0.0144 M solution. The pH of the resulting solution i
alexdok [17]

Answer:

The value of dissociation constant of the monoprotic acid is 1.099\times 10^{-3}.

Explanation:

The pH of the solution = 2.46

pH=-\log[H^+]

2.46=-\log[H^+]

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HA\rightleftharpoons H^++A^-

Initially

0.0144         0      0

At equilibrium

(0.0144-x)       x       x

The expression if an dissociation constant is given by :

K_a=\frac{[A^-][H^+]}{[HA]}

K_a=\frac{x\times x}{(0.0144-x)}

x=[H^+]=0.003467 M

K_a=\frac{0.003467 \times 0.003467 }{(0.0144-0.003467 )}

K_a=1.099\times 10^{-3}

The value of dissociation constant of the monoprotic acid is 1.099\times 10^{-3}.

3 0
4 years ago
The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe
Irina-Kira [14]

Answer:

t=147.24\ min

Explanation:

Given that:

Half life = 30 min

t_{1/2}=\frac{\ln2}{k}

Where, k is rate constant

So,  

k=\frac{\ln2}{t_{1/2}}

k=\frac{\ln2}{30}\ min^{-1}

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration [A_0] = 7.50 mg

Final concentration [A_t] = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

0.25=7.50e^{-0.0231\times t}

750e^{-0.0231t}=25

750e^{-0.0231t}=25

x=\frac{\ln \left(30\right)}{0.0231}

t=147.24\ min

6 0
4 years ago
The rate law for 2NO(g) + O₂(g) → 2NO₂(g) is rate = is rate k[NO]²[O₂]. is rateIn addition to the mechanism in the text (p. 709)
KonstantinChe [14]

The rate law depicts the effect of concentration on reaction rate. Second mechanism 2NO(g) ⇄ N₂O₂(g) [fast], N₂O₂(g) + O₂(g) → 2NO₂(g) [slow] is most reasonable. Thus, option b is correct.

<h3>What is rate law?</h3>

Rate law and equation give the rate at which the reaction takes place under the influence of the concentration of the reactants. The balanced chemical reaction is given as,

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The rate of the equation is given as,

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N₂O₂(g) +O₂(g) → 2NO₂ (g) [slow]

Rate is given as,

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Therefore, option b. the second mechanism is the most reasonable.

Learn more about rate law, here:

brainly.com/question/14779101

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5 0
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