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Virty [35]
2 years ago
15

The volume of tea in a typical teacup 0.2

Chemistry
1 answer:
kobusy [5.1K]2 years ago
3 0

Given :

Some statements.

To Find :

The SI units in each case.

Solution :

1) It is seconds which is denoted by seconds.

2) It is mass and units are kg. ( Not pounds because S.I unit is kg )

3) Since, we are talking about length is so its meter.

4) SI units of volume is given by meter³.

Hence, this is the required solution.

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The first step in the Ostwald process for producing nitric acid is 4NH3(g) + 5O2(g) --> 4NO(g) + 6H2O(g). If the reaction of
wel

Answer:

a. 77%

Explanation:

To solve the exercise, you need to know the limiting reagent. The limiting reagent is one that is consumed first in the chemical reaction. The other reagent is known as an excess reagent, and it is the one left over in a chemical reaction.

To know if ammonia or oxygen is the limiting reagent, we should relate the weights of both compounds in the reaction. We know that 4NH3 (68 g) react with 5O2 (160 g), so we should find out how much NH3 is required to react with 150 g of O2:

160 g O2 _______ 68 g NH3

150 g O2 _______ x = 150 g * 68 g / 160 g = 63.75 g NH3

As we have 150 g of NH3 but only 63.75 g are required, we know that ammonia is the excess reagent, so oxygen is the reagent that limits the reaction and that we should use to calculate the percentage yield of the reaction.

In the reaction we observe that 5O2 (160 g) produces 4NO (120g), so:

160 g O2 ______ 120 g NO

150 g O2 ______ x = 150 g * 120 g / 160 g = 112.5 g NO

If the percent yield of the reaction were 100%, it would produce 112.5 g of NO, however only 87 g of NO were obtained, so we should find out what the percentage of reaction yield was:

112.5 g NO ______ 100%

87 g NO _______ x = 87g * 100% / 112.5 g = 77%

Thus we observe that the reaction had a yield of 77%.

7 0
2 years ago
Read 2 more answers
When the elements in Group 1 are considered in order from top to bottom, each successive element at standard pressure has
kvasek [131]
When the element in group 1 are considered in order from top to bottom, each successive Elmer at standard pressure has a lower melting point and a lower boiling point.
7 0
2 years ago
Quinine, an antimalarial drug, is 8.63% nitrogen. There are two nitrogen atoms per molecule. What is the molecular weight of qui
Furkat [3]

Answer:

324.18 g/mol

Explanation:

Let the molecular mass of the antimalarial drug, Quinine is x g/mol

According to question,

Nitrogen present in the drug is 8.63% of x

So, mass of nitrogen = \frac {8.63}{100}\times x

Also, according to the question,

2 atoms are present in 1 molecule of the drug.

Mass of nitrogen = 14.01 amu = 14.01 g/mol (grams for 1 mole)

So, mass of nitrogen = 14.01×2 = 28.02

These 2 must be equal so,

\frac {8.63}{100}\times x=28.02

solving for x, we get:

<u>x = 324.18 g/mol</u>

6 0
2 years ago
How many milligrams of MgI2 must be added to 257.7 mL of 0.087 M KI to produce a solution with [I−] = 0.1000 M?
lbvjy [14]

Answer:

The answer is 465.6 mg of MgI₂ to be added.

Explanation:

We find the mole of ion I⁻ in the final solution

C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol

But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.

So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.

Hence, the weight of MgI₂ must be added is

Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg

4 0
2 years ago
6K + B2O3 → 3K2O + 2B
atroni [7]

Answer:

104.84 moles

Explanation:

Given data:

Moles of Boron produced = ?

Mass of B₂O₃ = 3650 g

Solution:

Chemical equation:

6K + B₂O₃    →    3K₂O + 2B

Number of moles of B₂O₃:

Number of moles = mass/ molar mass

Number of moles = 3650 g/ 69.63 g/mol

Number of moles = 52.42 mol

Now we will compare the moles of  B₂O₃ with B from balance chemical equation:

                 B₂O₃          :          B

                    1              :          2

                52.42         :        2×52.42 = 104.84

Thus from 3650 g of  B₂O₃  104.84 moles of boron will produced.

6 0
2 years ago
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