2500m into kilometer

Need help please With this.

Brums [2.3K]

Answer:question 3:(W)

question 4:(c)

Source:trust me bro

6 0

3 years ago

Read 2 more answers

67g of K2O is how many moles

taurus [48]

Follow the formula.. there’s not enough in this question for me to solve

7 0

3 years ago

A molecule of an organic compound contains at least one atom of

EleoNora [17]

Answer:

At least one atom contains an organic compound molecule. (1) sugars (3) sugar.

Explanation:

At least one atom contains an organic compound molecule. (1) sugars (3) sugar.

Answer is above

Hope this helps.

4 0

4 years ago

What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = pKa? Ka = 1.8×10-5

Westkost [7]

Answer:

a) We have to add 0.9 mole NaOH to 1.0 L of 1.8 M HC2H3O2

b) We have to add 0.277 mole NaOH to 1.0 L of 1.8 M HC2H3O2

c) We have to add 1.16 mole NaOH to 1.0 L of 1.8 M HC2H3O2

Explanation:

a) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = pKa?

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = pK = -log(1.8*10^-5) = 4.74

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

4.74 = 4.74 + log(A-/HA)

0 = log(A-/HA)

A-/HA = 1

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 1

X =0.9

We have to add 0.9 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

4.74 = 4.74 + log(0.9/0.9) = 4.74

b) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 4.00?

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = 4

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

4 = 4.74 + log(A-/HA)

-0.74 = log(A-/HA)

A-/HA = 0.182

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 0.182

X =0.277

We have to add 0.277 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

4 = 4.74 + log(0.277/1.523)

c) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 5.00

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = 5

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

5 = 4.74 + log(A-/HA)

0.26 = log(A-/HA)

A-/HA = 1.82

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 1.82

X =1.16

We have to add 1.16 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

5 = 4.74 + log(1.16/0.64) = 5

3 0

4 years ago

. Use the following reaction to determine how much of each product would be released if 42 000 kg (42 tonnes) of methyl isocyana

fenix001 [56]

The amount of 1,3-dimethyl urea produced would be 32,458 grams or 32.458 kg while that of carbon dioxide would be 16,214 grams of 16.214 kg

<h3>Stoichiometric problem</h3>

From the equation of the reaction, the mole ratio of methyl isocyanate with the products is 2:1 respectively.

Mole of 42,000 kg of methyl isocyanate = 42000/57 = 736.84 moles

Equivalent mole of 1,3-dimethyl urea = 736.84/2 =368.42 moles

Equivaent mole of carbon dioxide = 736.84/2 =368.42moles

Mass of 368.42 moles 1,3-dimethyl urea = 368.42 x 88.1 = 32,458 grams or 32.458 kg

Mass of 368.42 moles of carbon dioxide = 368.42 x 44.01 = 16,214 grams of 16.214 kg

More on stoichiometric problems can be found here: brainly.com/question/14465605

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5 0

2 years ago

2500m into kilometer​

2500m into kilometer