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3 years ago

Given the information that nitrogen atoms are larger than hydrogen atoms,which of the models best represent ammonia?(NH3)

Chemistry

1 answer:

Ray Of Light [21]3 years ago

4 0

Answer:

<em>The correct option is the structure in the option D. </em>

Explanation:

Ammonia contains a single nitrogen atom bonded to three hydrogen atoms.

The structure present in option A shows a single nitrogen atom bonded with a single hydrogen atom hence it is not correct.

The structure in option B shows a nitrogen atom bonded to four hydrogen atoms hence it is also not correct.

The structure in option C shows a nitrogen atom bonded to two hydrogen atoms hence it is also not correct.

The structure in option D shows a nitrogen atom bonded to three hydrogen atoms which is the correct structure of ammonia (NH3).

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Five students performed a Kjeldahl nitrogen analysis of a protein sample. The following weight % nitrogen values were determined

Sunny_sXe [5.5K]

Answer:

G_calculated = 1.756

The outlier should be rejected, as G_cal > G_tab (= 1.463) at 95 % confidence.

Explanation:

The Grubb's test is used for identifying an outlier in data, which is from the same population. For this, a statistical term, G, is calculated for the suspected outlier. If the calculated value is greater than the tabulated G value then the suspected value is rejected. This term is given as,

G_calculated = | suspect value - mean| / s

Here, suspect value is 13.8, mean is to be taken of all the data (including suspected value). s is the standard deviation of the sample data.

s is calculated from the following formula:

s = (Σ(xi - x)²/(N-1))^1/2

Here, x is the mean, which is 15.24, xi is individual value and N is the total number of data (5).

From the above formula, s is found to be

Standard Deviation, s = 0.820

Now for G value,

G_calculated = | 13.8 - 15.24| / (0.820)

G_ calculated = 1.756

The tabulated G value at 95 % confidence and N -1 (5 - 1 = 4) degree of freedom is, 1.463.

As calculated G (1.756) is greater than the tabulated G (1.463), the value 13.8 is considered an outlier at 95 % confidence.

3 0

3 years ago

Which of the following is the best choice for presenting the percent

Alik [6]

OD

hope this helps!!

3 0

3 years ago

Read 2 more answers

Please help!

Dahasolnce [82]

Answer:

reproduction

Explanation:

reproduction, process by which organisms replicate themselves

6 0

3 years ago

If 45 mL of water are added to 250 mL of a 0.75 M K2SO4 solution, what will the molarity of the diluted solution be?

krok68 [10]

Answer:

$\large\boxed{\large\boxed{0.64M}}$

Explanation:

When you form a <em>diluted solution</em> from a mother (concentrated) solution, the moles of solute are determined by the mother solution.

The main equation is:

$Molarity=\dfrac{\text{moles of solute}}{\text{volume of the solution in liters}}$

Then, since the moles of solute is the same for both the mother solution and the diluted solution:

$\text{Molarity mother solution }\times\text{ volume mother solution}=\\\\=\text{Molarity diluted solution }\times\text{ volume diluted solution}$

Substitute and solve for the molarity of the diluted solution:

$250mL\times 0.75M=(45mL+250mL)\times M\\\\\\M=\dfrac{250mL\times 0.75M}{295mL}=0.64M$

8 0

3 years ago

2. Fe(s) + O2(g) → Fe3O4(s) a. When 13.54 g of O2 is mixed with 12.21 g of Fe, which is the limiting reactant? b. What mass in g

julsineya [31]

The percent yield shows the extent to which the reactants are converetd into products . The limiting reactant is used up in the reaction.

<h3>What is a limiting reactant?</h3>

A limiting reactant is the reactant that is in the least amount in the system. Now;

Number of moles of Fe = 12.21 g/56 g/mol = 0.22 moles

Number of moles of O2 = 13.54 g/32 g/mol = 0.42 moles

Balanced reaction equation;

3Fe(s) + 2 O2(g) = Fe3O4(s)

If 3 moles of Fe reacts with 2 mole of O2

0.22 moles of Fe reacts with 0.22 moles * 2 mole/3 moles = 0.15 moles

Hence, Fe is the limiting reactant

If 3 mole of Fe produces 1 mole of Fe3O4(s)

0.22 moles of O2 produces 0.22 moles * 1 mole/3 moles of Fe3O4(s) = 0.1073 moles

Mass of Fe3O4(s) =0.1073 moles * 232 g/mol =16.9 g

Number of moles of excess reactant = 0.42 moles - 0.15 moles = 0.27 moles

Mass of excess reactant = 0.27 moles * 32 g/mol = 8.64 g

percent yield = 15.88 g/16.9 g * 100/1

= 93.4%

Learn more about percent yiled: brainly.com/question/13463225

8 0

2 years ago