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2 years ago

0.475 g H, 7.557 gS, 15.107 g O. Express your answer as a chemical formula.

Chemistry

1 answer:

max2010maxim [7]2 years ago

7 0

Answer:

H₂SO₄

Explanation:

We have a compound formed by 0.475 g H, 7.557 g S, 15.107 g O. In order to determine the empirical formula, we have to follow a series of steps.

Step 1: Calculate the total mass of the compound

Total mass = mass H + mass S + mass O = 0.475 g + 7.557 g + 15.107 g

Total mass = 23.139 g

Step 2: Determine the percent composition.

H: (0.475g/23.139g) × 100% = 2.05%

S: (7.557g/23.139g) × 100% = 32.66%

O: (15.107g/23.139g) × 100% = 65.29%

Step 3: Divide each percentage by the atomic mass of the element

H: 2.05/1.01 = 2.03

S: 32.66/32.07 = 1.018

O: 65.29/16.00 = 4.081

Step 4: Divide all the numbers by the smallest one

H: 2.03/1.018 ≈ 2

S: 1.018/1.018 = 1

O: 4.081/1.018 ≈ 4

The empirical formula of the compound is H₂SO₄.

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2 years ago

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2 years ago

The enthalpy change for converting 1.00 mol of ice at -25.0 ∘c to water at 90.0∘c is ________ kj. the specific heats of ice, wat

liubo4ka [24]

Answer : The enthalpy change for converting 1 mole of ice at $-25.0^oC$ to water at $90^oC$ is, 7.712 KJ

Solution :

Process involved in the calculation of enthalpy change :

$(1):ice(-25^oC)\rightarrow ice(0^oC)\\\\(2):ice(0^oC)\rightarrow water(0^oC)\\\\(3):water(0^oC)\rightarrow water(90^oC)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{ice}\times (T_2-T_1)]+\Delta H_{fusion}+[m\times c_{water}\times (T_3-T_2)]$

where,

$\Delta H$ = enthalpy change

m = mass of water = $1mole\times 18g/mole=18g$

$c_{ice}$ = specific heat of ice = 2.09 J/gk

$c_{water}$ = specific heat of water = 4.18 J/gk

$\Delta H_{fusion}$ = enthalpy change for fusion = 6.01 KJ/mole = 0.00601 J/mole

conversion : $0^oC=273k$

$T_1$ = initial temperature of ice = $0^oC=273k$

$T_2$ = final temperature of ice = $-25^oC=273+(-25)=248k$

$T_3$ = initial temperature of water = $0^oC=273k$

$T_4$ = final temperature of water = $90^oC=273+90=363k$

Now put all the given values in the above expression, we get

$\Delta H=[18g\times 2.09J/gK\times (273-248)k]+0.00601J+[18g\times 4.18J/gK\times (363-273)k]$

$\Delta H=7712.106J=7.712KJ$ (1 KJ = 1000 J)

Therefore, the enthalpy change for converting 1 mole of ice at $-25.0^oC$ to water at $90^oC$ is, 7.712 KJ

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2 years ago

You have just started working at a swimming pool supply store and they have nitric acid on their shelves. In order to help custo

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Answer:

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Explanation:

The aqueous solution ionization reaction for nitric acid is as follows;

HNO₃ + H₂O → NO₃⁻ + H₃O⁺

Whereby HNO₃ is a strong acid and it undergoes complete ionization in an aqueous solution

Hence one mole of HNO₃ produces one mole of H₃O⁺ ions

0.0155 mol/L HNO₃ will have 0.0155 moles of H₃O⁺ in one liter of the solution

The concentration of hydronium ions = 0.0155 mol/L.

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2 years ago

A student takes a measured volume of 3.00 M HCl to prepare a 50.0 mL sample of 1.80 M HCl. What volume of 3.00 M HCl did the stu

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Explanation:

M $_{i}$ = 3M

V $_{i}$ = 50mL

M $_{f}$ = 1.80M

Unknown:

V $_{f}$ = ?

Solution:

To solve this problem we must understand that the initial and final number of moles remains the same.

Number of moles = molarity x volume

Molarity x volume of initial solution = Molarity x volume of final solution

Therefore:

M $_{i}$ x V $_{i}$ = M $_{f}$ x V $_{f}$

M is the molarity

V is the volume

i = initial state

f = final state

input the parameters;

3 x 50 = 1.8 x V $_{f}$

V $_{f}$ = 83.3 mL

learn more:

Molarity brainly.com/question/9324116

#learnwithBrainly

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2 years ago

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