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Tcecarenko [31]
3 years ago
8

When sulfur trioxide gas reacts with water, a solution of sulfuric acid forms.Express your answer as a balanced chemical equatio

n. Identify all of the phases in your answer.
Chemistry
1 answer:
Jlenok [28]3 years ago
8 0

Answer: SO_3(g)+H_2O(l)\rightarrow H_2SO_4(aq)

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The phases are represented as (s) for solid sate, (l) for liquid state, (g) for gaseous state and (aq) for aqueous state.

Thus the balanced chemical equation for :sulfur trioxide gas reacts with water, a solution of sulfuric acid forms is :

SO_3(g)+H_2O(l)\rightarrow H_2SO_4(aq)

You might be interested in
How many ions are present in 75.0g of magnesium sulfate
Ostrovityanka [42]

Answer:

7.5×10²³ ions

Explanation:

Given data:

Mass of magnesium sulfate = 75.0 g

Number of ions = ?

Solution:

Number of moles of magnesium sulfate:

Number of moles = mass/molar mass

Number of moles = 75.0 g/ 120.36 g/mol

Number of moles = 0.623 mol

Magnesium sulfate formula: MgSO₄

Number of moles of ions in 1 mole of MgSO₄ = 2 mol (Mg²⁺ and SO₄²⁻)

Number of moles of ion in 0.623 moles of MgSO₄:

2 mole ×0.623 = 1.246 mol

Number of ions:

1 mole contain 6.022×10²³ ions

1.246 mol × 6.022×10²³ ions / 1mol

7.5×10²³ ions

4 0
3 years ago
What is the mass of one mole of S8?​
vredina [299]

Answer:

256.52 g/m

Explanation:

Each S has a molecular mass of 32.056 g/mol so multiply that by 8 and you get 256.52.

4 0
3 years ago
How many moles of each element are in one mole of (NH4)2S? (3 points)
cricket20 [7]

Answer:

b 1 mole of nitrogen 6 hydro 1 sulfur

3 0
3 years ago
Read 2 more answers
If Steve throws a football 57 meters in 3 seconds, what is the average speed of the football?​
sattari [20]

Answer:

19 m/s

Explanation:

57/3

7 0
2 years ago
A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
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