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makkiz [27]
3 years ago
7

hat is the percentage of water in the following compound? Answer using three significant figures. Sodium carbonate decahydrate,

Na2CO3 • 10H2O
Chemistry
1 answer:
BigorU [14]3 years ago
7 0
You need to find the whole molar mass of the compound using the periodic table to add the values. 

Na2CO3= (2 x 23.0) + 12.0 + (3 x 16.0)= 106 g/mol
H2O= 10 x [ (2 x 1.01 ) + (16.0) ]= 180.2 g/mol

the total molar mass is 106 + 180.2 = 286.2 g/mol

the percentage of water you can find by doing "parts over the whole"

H2O%= 180.2 / 286.2 X 100= 63.0%
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Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
How does matter change from one state to another?
Rzqust [24]
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Bas +<br> PtF2 →<br> BaF2 +<br> Pts<br> Need to balance it
Basile [38]

it is already balanced

REACTANTS

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PRODUCT

Barium fluoride (BaF2) + Cooperite (PtS)

Hope this answer helps you dear! take care

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