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2 years ago

I need help with this please

Chemistry

1 answer:

Annette [7]2 years ago

8 0

Answer:

1=Rb, 2=O, 3=As, 4=Al and 5=Ca

Explanation:

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The sun is 150 kilometers from earth.if you wanted to explain to friend how far is what would you say

jasenka [17]

Answer:

150 times 1000 is how many meters it is away from earth because kilo means 1000

Explanation:

easy metric conversions

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3 years ago

Which system includes feet, pounds, gallons, and miles? metric customary

slavikrds [6]

Answer:

customary Is the required system

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2 years ago

The acid HOCl (hypochlorous acid) is produced by bubbling chlorine gas through a suspension of solid mercury(II) oxide particles

sergejj [24]

Answer: The expression for equilibrium constant is $K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}$

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{eq}$

For the general chemical equation:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_c$ is given as:

$K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

For the given chemical reaction:

$2HgO(s)+H_2O(l)+2Cl_2(g)\rightleftharpoons 2HOCl(aq.)+HgO.HgCl_2(s)$

The expression for $K_{eq}$ is given as:

$K_{eq}=\frac{[HOCl]^2[HgO.HgCl_2]}{[HgO]^2[H_2O][Cl_2]^2}$

The concentration of solid is taken to be 0.

So, the expression for $K_{eq}$ is given as:

$K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}$

3 0

3 years ago

Choose the products that complete the reaction . The chemical equation may not be balanced . Al+H 2 SO 4 ? Al 2 (SO 4 ) 3 +H 2 O

alexandr402 [8]

Answer: Al2 (SO4)3 + H2

Explanation:

option C

5 0

2 years ago

A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita

mestny [16]

Answer:

The answer is:

(a) $NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)$

(b) NaCl

(c) 0.211 g

Explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of $AgNO_3$ ,

= 30.0 ml

or,

= 0.030 L

Molarity of $AgNO_3$ ,

= 0.050 M

Moles of NaCl will be:

= $\frac{Given \ mass}{Molar \ mass}$

= $\frac{0.0860}{58.44}$

= $0.00147 \ mol$

now,

Moles of $AgNO_3$ will be:

$= Molarity\times Volume$

$= 0.050\times 0.030$

$=0.0015 \ mol$

(a)

The reaction is:

⇒ $NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)$

(b)

1 mole of NaCl react with,

= 1 mol of $AgNO_3$

0.0015 mol $AgNO_3$ needs,

= $0.00150 \ mol \ NaCl$

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

= $0.00147\times \frac{1}{1}\times \frac{143.32}{1}$

= $0.211 \ g \ AgCl$

4 0

3 years ago