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IrinaK [193]
2 years ago
10

SOMEONE HELP NO LINKS PLZ answer Questions 1–6 based on Figure 1.

Chemistry
1 answer:
Bogdan [553]2 years ago
6 0

Answer:

element argon, protons 18 , electrons 18 , outer most energy level 8

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Use the following balanced equation to answer the questions below.
Blababa [14]

Answer:

A. 4.5 mol Mg(OH)₂

B. 6 mol NaOH

Explanation:

Let's consider the following balanced equation.

Mg(NO₃)₂ + 2 NaOH ⇒ Mg(OH)₂ + 2 NaNO₃

PART A

The molar ratio of NaOH to Mg(OH)₂ is 2:1. The moles of Mg(OH)₂ produced from 9 moles of NaOH are:

9 mol NaOH × 1 mol Mg(OH)₂/2 mol NaOH = 4.5 mol Mg(OH)₂

PART B

The molar ratio of NaOH to NaNO₃ is 2:2. The moles of NaOH needed to produce 6 moles of NaNO₃ are:

6 mol NaNO₃ × 2 mol NaOH/2 mol NaNO₃ = 6 mol NaOH

5 0
3 years ago
explain why ionic compounds are not electrical conductors in the solid state, but conduct electricity when dissolved.
Alik [6]

Answer:

Hi !

Here is your answer !

Ionic compounds cannot conduct electricity when solid, as their ions are held in fixed positions and cannot move.

Explanation:

Ionic compounds conduct electricity when molten (liquid) or in aqueous solution (dissolved in water), because their ions are free to move from place to place.

Thank You !

4 0
3 years ago
Lily took 57 seconds to walk from classroom to library. If the distance between the classroom and library was 38 m, at what aver
slamgirl [31]
1.5 sorry if I’m wrong!
4 0
3 years ago
If u answer this correctly I’ll mark you brainliest
olga2289 [7]

Answer:

6 is the right answer I know cause I like science

4 0
3 years ago
Read 2 more answers
Consider the following reaction, equilibrium concentrations, and equilibrium constant at
REY [17]

Answer:

Equilibrium concentration of H_{2}O is 12.5 M

Explanation:

Given reaction: C_{2}H_{4}+H_{2}O\rightleftharpoons C_{2}H_{5}OH

Here, K_{c}=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}][H_{2}O]}

where K_{c} represents equilibrium constant in terms of concentration and species inside third bracket represent equilibrium concentrations

Here, [C_{2}H_{4}]=0.015M , [C_{2}H_{5}OH]=1.69M and K_{c}=9.0

So, [H_{2}O]=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}]\times K_{c}}=\frac{1.69}{0.015\times 9.0}=12.5M

Hence equilibrium concentration of H_{2}O is 12.5 M

5 0
3 years ago
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