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Lelechka [254]
3 years ago
10

What is the golden rule of lighting safety?

Chemistry
1 answer:
frozen [14]3 years ago
4 0

The golden rule is to head for cover.

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Can someone help me on 1 and 2
pav-90 [236]
I believe 1 is growth and 2 is reproduction. Hope this helps.
6 0
3 years ago
Read 2 more answers
Predict the products of La(s) + O2(aq) ->
mina [271]

Answer:

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

Explanation:

<em>Hello </em><em>there!</em>

When you are given such a problem for completing the chemical equations, what you have to understand is that metals are found in groups I, II and III. While Oxygen is a group VI element.

From the above question I have considered that my La(s), solid is either Sodium (Na) - group I, Magnesium - group II and Aluminum - group III.

In a reaction, there is exchange of electrons given by their oxidation numbers (I, II and III - for our metals above)

The chemical equations are thus;

i. Na (s) + O2 (g) => NaO (s)

ii. Mg (s) + O2 (g) => Mg2O (s)

iii. Al (s) + O2 (g) => Al2O3 (s)

Relate this to the problem and it will be;

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

<em>I hope this </em><em>helps </em><em>you</em><em> </em><em>to </em><em>understand</em><em> </em><em>better</em><em>.</em><em> </em><em>Enjoy </em><em>your</em><em> </em><em>studies</em>

3 0
2 years ago
Two unknown molecular compounds were being studied. A solution containing 5.00 g of compound A in 100. g of water froze at a low
LenaWriter [7]

Answer:

Compound B has greater molar mass.

Explanation:

The depression in freezing point is given by ;

\Delta T_f=i\times k_f\times m..[1]

m=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}

Where:

i = van't Hoff factor

k_f = Molal depression constant

m = molality of the solution

According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.

The depression in freezing point of solution with A solute: \Delta T_{f,A}

Molar mass of A = M_A

The depression in freezing point of solution with B solute: \Delta T_{f,B}

Molar mass of B = M_B

\Delta T_{f,A}>\Delta T_{f,B}

As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.

\Delta T_f\propto \frac{1}{\text{Molar mass of solute}}

M_A

This means compound B has greater molar mass than compound A,

4 0
3 years ago
The student's lab manual says to mix some of his Na2CO3 solution with an aqueous solution of copper(II) sulfate (CuSO4)
lord [1]

Explanation:

When the student mixed the solution sodium carbonate with solution of copper(II) sulfate ; Copper Hydroxocarbonate , sodium sulfate and carbon dioxide gas was obtained as a products.

The balanced chemical reaction

2Na_2CO_3+2CuSO_4\rightarrow Cu_2(OH)_2CO_3+2Na_2SO_4+CO_2

Where:

Cu_2(OH)_2CO_3 = Copper(II) Hydroxocarbonate

Na_2CO_3 = Sodium carbonate

CuSO_4 = Copper(II) sulfate

Na_2SO_4 = Sodium sulfate

CO_2 = Carbon-dioxide

3 0
3 years ago
What is the concentration of each ion in a solution that is prepared by dissolving 5.00 g of ammonium chloride in enough water t
const2013 [10]

Answer:

C = 0.08M

Explanation:

molar mass of AlCl3

Al =27

Cl = 35.5

27+3(35.5) =133.5g/mol

n= mass/Molar mass

n =CV

CV = mass/molar mass

C x 500 x 10^-³ = 5/133.5

C x 500 x 10^-³ = 0.04

C = 0.04/500 x 10^-³

C = 0.08M

4 0
3 years ago
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