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3 years ago

Point N is on line segment MO. Given NO = 2x – 3, MO = 3x + 5, and

Chemistry

1 answer:

Zigmanuir [339]3 years ago

5 0

Answer:

MO = 20

Explanation:

Given that: MO = 3x + 5

MN = 2x + 3

NO = 2x - 3

But,

MO = MN + NO

So that;

3x + 5 = (2x + 3) + (2x - 3)

3x + 5 = 2x + 3 + 2x - 3

3x + 5 = 4x

3x - 4x = -5

-x = -5

Multiply through by minus,

x = 5

Thus,

MO = 3x + 5

= 3(5) + 5

= 15 + 5

MO = 20

The length of MO is 20.

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Calculate the number of moles in 2.67 g of carbon trihydride

Scorpion4ik [409]

Answer:0.178 moles

Explanation: carbon trihydride seems to be an unusual name for the methyl group CH3–

ionic wt 15

moles = 2.67/15 = 0.178

8 0

3 years ago

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant o

sattari [20]

Answer:

4.26 %

Explanation:

There is some info missing. I think this is the original question.

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10 ⁻⁴.

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50 × 10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

$[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M$

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

$\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%$

4 0

4 years ago

If not managed properly, aquaculture can

Damm [24]

C. pollute water and damage aquatic ecosystems

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3 years ago

Can you please help

Digiron [165]

Answer:

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3 years ago

Using the reaction below determine the amount of Sulfur proceduced in grams (this is a limiting reaction question) 2 H2S + SO2 -

marusya05 [52]

Answer:

= 2.2 g pf S. produced

Explanation:

Balanced Reaction equation:

$2H_{2} S + SO_{2}$ → $3S + 2H_{2} O$

1 mole of H2S - 34.1g

? moles - 3.2g

= 3.2/34.1 = 0.09 moles of H2S

Also,

1 mole of S02 - 64.07 g

? moles - 4.42g

= 4.42/64.07 = 0.069 moles of SO2

Meaning SO2 is the limiting reagent

Finally, 3 moles of S - 32g of sulphur

0.069 mole = ? g of Sulphur

= 0.069 x 32

= 2.2 g pf S.

7 0

3 years ago