Which of the following is true regarding atomic radius?

20.0 g of bromic acid, HBrO3, is reacted with excess HBr.

Blababa [14]

After Rounding off The percentage yield is 64%

<h3>What is Percentage Yield ?</h3>

It is the ratio of actual yield to theoretical yield multiplied by 100% .

It is given in the question

20.0 g of bromic acid, HBrO3, is reacted with excess HBr.

The reaction is

HBrO₃ (aq) + 5 HBr (aq) → 3 H₂O (l) + 3 Br₂ (aq)

Actual yield = 47.3 grams

Molecular weight of Bromic Acid is 128.91 gram

Moles of Bromic Acid = 20/128.91 = 0.155 mole

Mole fraction ratio of Bromic Acid to Bromine is 1 :3

Therefore for 0.155 mole of Bromic Acid 3 * 0.155 = 0.465 mole of Bromine is produced.

1 mole of Bromine = 159.8 grams of Bromine

0.465 of Bromine = 74.31 grams of Bromine

Percentage Yield = (47.3/74.31)*100 = 63.65 %

After Rounding off The percentage yield is 64% .

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5 0

2 years ago

Under standard-state conditions, which of the following half-reactions occurs at the cathode during the electrolysis of aqueous

Slav-nsk [51]

Answer:

The following reaction will occur at cathode:

$Ni^{+2}(aq)+2e--->Ni(s)$

Explanation:

The two half reaction during electrolysis of aqueous nickel sulfate will be

a) anode reaction :

Water will undergo oxidation and will evolve oxygen gas at anode as shown in the given reaction:

$2H_{2}O(l) ----> O_{2}(g) + 4H^{+}(aq) + 4e$

b) Cathode reaction: The reduction of Nickel ion will occur by gain of two electrons as shown in the given equation:

$Ni^{+2}(aq)+2e--->Ni(s)$

Thus the overall reaction will be:

$2H_{2}O(l) + 2Ni^{+2}----> O_{2}(g) + 4H^{+}(aq) + 2Ni(s)$

6 0

3 years ago

Determine whether each of the following cations is acidic or pH-neutral.

riadik2000 [5.3K]

Answer:

1. Acidic

2.Acidic

3, Acidic

All are acidic

5 0

3 years ago

Two workers used a force of 100 N to move a sofa 4 meters in 2 seconds how much power

r-ruslan [8.4K]

Answer:

200 watts

Explanation:

First we have to find total work:

W =F(d)

W = 400J

Then plug that answer into the Power formula.

P = W/(t_f-t-i)

P = 400/2s

P = 200 W

3 0

3 years ago

Benzene is a minor component of gasoline. The standard molar enthalpy of formation of benzene C7H16(l) is 48.95 kJ/mol. For the

Radda [10]

Answer:

-3135.47 kJ/mol

Explanation:

Step 1: Write the balanced equation

C₆H₆(l) + 7.5 O₂(g) ⇒ 6 CO₂(g) + 3 H₂O(g)

Step 2: Calculate the standard enthalpy change of the reaction (ΔH°r)

We will use the following expression.

ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpies of formation

p: products

r: reactants

ΔH°r = 6 mol × ΔH°f(CO₂(g)) + 3 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C₆H₆(l)) - 7.5 mol × ΔH°f(O₂(g))

ΔH°r = 6 mol × (-393.51 kJ/mol) + 3 mol × (-241.82 kJ/mol) - 1 mol × (48.95 kJ/mol) - 7.5 mol × 0 kJ/mol

ΔH°r = -3135.47 kJ

Since this enthalpy change is for 1 mole of C₆H₆(l), we can express it as -3135.47 kJ/mol.

7 0

3 years ago