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3 years ago

How many elements are in a set whole numbers between 3 and 15

1 answer:

3 0

The set of elements composed of whole numbers between 3 and 15 consists of 4, 5, 6,7,8,9, 10, 11,12,13,14. Counting these, the number of elements is 11.

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Ferrous sulfate (FeSO4) was used in the

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Ferrous sulfate (FeSO4) is obtained by the reaction of aqueous ferric sulfate with gaseous sulfur dioxide and water according to the equation below:

Aqueous ferric sulfate = $Fe_2(SO_4)_3_{(aq)}$

Gaseous sulfur dioxide = $SO_2_{(g)}$

Water = $H_2O_{(l)}$

Aqueous sulfuric acid = $H_2SO_4_{ (aq)}$

Aqueous ferrous sulfate = $FeSO_4_{ (aq)}$

Thus, the balance equation becomes:

$Fe_2(SO_4)_3_{(aq)}$ + $SO_2_{(g)}$ + $2H_2O_{(l)}$ $--->$ $2 H_2SO_4_{ (aq)} + 2 FeSO_4_{ (aq)}$

More on ferrous sulfate can be found here: brainly.com/question/10834028

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Answer: limiting reactant controls the amount of product formed in a chemical reaction.

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4 years ago

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Read 2 more answers

3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00$

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

$n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00 \approx 5$

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

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4 years ago