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3 years ago

How something works is related to its structure

Chemistry

1 answer:

boyakko [2]3 years ago

6 0

True?

Step by step explanation:
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1. Starting with 9.3 moles of O 2 , how many moles of H 2 S will be needed and how many moles of SO 2 will

enyata [817]

Answer:

Moles of H₂S needed = 6.2 mol

Moles of SO₂ produced = 6.2 mol

Explanation:

Given data:

Number of moles of O₂ = 9.3 mol

Moles of H₂S needed = ?

Moles of SO₂ produced = ?

Solution:

Chemical equation:

2H₂S + 3O₂ → 2SO₂ + 2H₂O

Now we will compare the moles of oxygen with H₂S.

O₂ : H₂S

3 : 2

9.3 : 2/3×9.3 = 6.2 mol

Now we will compare the moles of SO₂ with both reactant.

O₂ : SO₂

3 : 2

9.3 : 2/3×9.3 = 6.2 mol

H₂S : SO₂

2 : 2

6.2 : 6.2 mol

So 6.2 moles of SO₂ are produced.

6 0

3 years ago

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$

calculating the reverse reaction

$H_2(g) + I_2(g)\rightleftharpoons 2HI(g)$

$Kc = \frac{1}{Kc} \\\\$

$= \frac{1}{0.034386}\\ \\= 29.081\\$

7 0

3 years ago

7. What is the voltage when the resistance is 6 ohms and the current is 8 amps?

vivado [14]

Answer:

48 volts

Explanation:

Voltage (E) = Current (I) x Resistance (R), or E = IR.

4 0

3 years ago

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Review this reaction:H2SO4 + NaOH ?.What are the product(s)?Na2SO4H2ONa2SO4 and H2ONaH and SO4OHH3SO5

muminat

i think its MIDDLE FINGERS UP IN THE SKY AND AT THESE AHOLE MODERATORS

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3 years ago

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Please help!<br> I don't understand, I asked for help but no one knows the answer.

faltersainse [42]

I've put in some of the answers.
hope this helps :)

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3 years ago

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