Answer:
6. O₂ + Cu —> CuO
7. H₂ + Fe₂O₃ —> H₂O + Fe
8. O₂ + H₂ — > H₂O
9. H₂S + NaOH —> Na₂S + H₂O
10. Al + HCl —> H₂ + AlCl₃
Explanation:
6. Oxygen gas react with solid copper metal to form copper(II) oxide
Oxygen gas => O₂
Copper => Cu
copper(II) oxide => CuO
The equation is:
O₂ + Cu —> CuO
7. hydrogen gas and iron(III) oxide powder react to form liquid water and solid iron power
hydrogen gas => H₂
Iron(III) oxide => Fe₂O₃
Water => H₂O
Iron => Fe
The equation is:
H₂ + Fe₂O₃ —> H₂O + Fe
8. Oxygen gas react with hydrogen gas to form liquid water
Oxygen gas => O₂
hydrogen gas => H₂
Water => H₂O
The equation is:
O₂ + H₂ — > H₂O
9. Hydrogen sulphide gas is bubbled through a sodium hydroxide solution to produce sodium sulphide and liquid water
hydrogen sulphide => H₂S
sodium hydroxide => NaOH
Sodium sulphide => Na₂S
Water => H₂O
The equation is:
H₂S + NaOH —> Na₂S + H₂O
10. Hydrogen gas and aluminum chloride solutions are produced when solid aluminum react with hydrochloric acid
Aluminum => Al
Hydrochloric acid => HCl
hydrogen gas => H₂
Aluminum chloride => AlCl₃
The equation is:
Al + HCl —> H₂ + AlCl₃
When y equals 5, x is 104.3
When y equals 3 then x is 108.3
<em><u>Solution:</u></em>
<em><u>Given expression is:</u></em>
<h3><u>If y equals 5 what is x ?</u></h3>
Substitute y = 5 in given expression
5 = 57.15 - 0.5(x)
5 = 57.15 - 0.5x
0.5x = 57.15 - 5
0.5x = 52.15
Divide both sides by 0.5
x = 104.3
Thus when y equals 5, x is 104.3
<h3><u>If y = 3 what is x ?</u></h3>
Substitute y = 3 in given expression
3 = 57.15 - 0.5(x)
3 = 57.15 - 0.5x
0.5x = 57.15 - 3
0.5x = 54.15
Divide both sides by 0.5
x = 108.3
Thus when y equals 3 then x is 108.3
The molar concentration of the nitric acid solution was 0.6666 mol/L.
<em>Balanced equation</em>: KOH + HNO_3 → KNO_3 + H_2O
<em>Moles of KOH</em>: 32.33 mL KOH × (1.031 mmol KOH /1 mL KOH)
= 33.33 mmol KOH
<em>Moles of HNO_3</em>: 33.33 mmol KOH× (1 mmol HNO_3/1 mmol KOH)
= 33.33 mmol HNO_3
<em>Concentration of KOH</em>: <em>c </em>= "moles"/"litres" = 33.33 mmol/50.00 mL
= 0.6666 mol/L
Answer:
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