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professor190 [17]
3 years ago
6

Write out the balanced equation for the following word equation. Use the proper state abbreviations for each of the reactants an

d products as well:
Lithium metal reacts with liquid water to form hydrogen gas and aqueous lithium hydroxide.
Chemistry
1 answer:
skad [1K]3 years ago
3 0

Answer:

Li (s)  +   H₂O (l)  → 1/2 H₂ (g)  + LiOH (aq)

Explanation:

Lithium metal reacts with liquid water to form hydrogen gas and aqueous lithium hydroxide

2Li (s)  +  2H₂O (l)  → H₂ (g)  + 2LiOH (aq)

We can divide stoichiometry coefficients /2, to have the lowest stoichiometry.

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3 years ago
Question 1
yanalaym [24]

Answer:

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Explanation:

6 0
3 years ago
The air inside a flexible 3.5L container has a pressure of 115Kpa. What should the volume of the container be increased to in th
LenaWriter [7]

Answer:

V₂ = 4.82 L

Explanation:

Given data:

Initial volume of gas = 3.5 L

Initial pressure = 115 Kpa

Final volume = ?

Final Pressure = 625 torr

Solution:

Final Pressure = 625 torr (625/760 =0.82 atm)

Initial pressure = 115 Kpa (115/101 = 1.13 atm)

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

1.13 atm × 3.5 L = 0.82 atm × V₂

V₂ = 3.955 atm. L/0.82 atm

V₂ = 4.82 L

5 0
4 years ago
PLEASE HELP- I WILL GIVE BRAINLIEST AND THANKS
julsineya [31]

Answer:

The mass of 10 cm³of a 0.4 g/dm³ solution of sodium carbonate is 0.004 grams

Explanation:

The question is with regards to density calculations

The density of the given sodium carbonate solution, ρ = 0.4 g/dm³

The volume of the given solution of sodium carbonate, V = 10 cm³ = 0.01 dm³

Density \ of \ an \ object, \rho  = \dfrac{The \ mass \ of \ the \ object, \ m }{\ The \ volume \ of \ the \ object, \ V }

\rho = \dfrac{m}{V}

Therefore, we have;

The \ Density \ of \ the \ sodium \ carbonate, \ \rho  = 0.4 \ g/dm^3 =  \dfrac{m }{ 0.01 \ dm^3 }

The mass, "m", of the sodium carbonate in  = ρ×V = 0.4 g/dm³ × 0.01 dm³ = 0.004 g

The mass of 10 cm³ (10 cm³ = 0.01 dm³) of a 0.4 g/dm³ solution of sodium carbonate, m = 0.004 g.

8 0
3 years ago
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