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2 years ago

If 13 grams of copper sulfate is reacted with zinc how much of each product is produced?

Chemistry

1 answer:

Sladkaya [172]2 years ago

7 0

Answer:

No. Of Moles of zinc = m/Ar

= 13/ 65.38 = 0.198 moles

From balanced equation, Mole ration between CuSO4 and Zn is 1 : 1

So only 0.198 moles of CuSO4 reacts, it is in excess

Mass = no of Moles X Mr

Mass = 0.198 X 159.5 = 31.59 grams

Volume = mass m denisty

Volume j 31.59 / 3.6 = 8.78 ml

Explanation:

i think this wrong

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If the molar absorptivity constant for the red dye solution is 5.56×104 M-1cm-1, calculate the molarity of the red dye solution

Shtirlitz [24]

Explanation:

a) Using Beer-Lambert's law :

Formula used :

$A=\epsilon \times c\times l$

where,

A = absorbance of solution = 0.945

c = concentration of solution = ?

l = length of the cell = 1.20 cm

$\epsilon$ = molar absorptivity of this solution = $5.56\times 10^4 M^{-1} cm^{-1}$

$0.945=5.56\times 10^4 M^{-1} cm^{-1}\times 1.20 \times c$

$c=1.4163\times 10^{-5} M=14.16 \mu M$

( $1\mu M=10^{-6} M$ )

14.16 μM is the molarity of the red dye solution at the optimal wavelength 519nm and absorbance value 0.945.

b) $c=1.4163\times 10^{-5} mol/L$

1 L of solution contains $1.4163\times 10^{-5}$ moles of red dye.

Mass of $1.4163\times 10^{-5}$ moles of red dye:

$1.4163\times 10^{-5}\times 879.86g/mol=0.01246 g$

$(w/v)\%=\frac{\text{Mass of solute (g)}}{\text{Volume of solvent (mL)}}\times 100$

$red(w/v)\%=\frac{0.01246 g}{1000 mL}\times 100=0.001246\%$

c) In order to dilute red dye solution by 5 times, we will need to add 1 L of water to solution of given concentration.

Concentration of red dye solution = $c=1.4163\times 10^{-5} M$

Concentration of red solution after dilution = c'

$c=c'\times 5$

$1.4163\times 10^{-5} M=c'\times 5$

$c'=2.83\times 10^{-6} M$

The final concentration of the diluted solution is $2.83\times 10^{-6} M$

8 0

3 years ago

(a) Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. (b) S

Sunny_sXe [5.5K]

These are two questions and two answers

Answer:

Question 1:

H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l)

Question 2:

0.201 M

Explanation:

Question 1:

The neutralization reaction that occurs between H₂SO₄ and KOH is an acid-base reaction.

The products of an acid-base reaction are salt and water.

This is the sketch of such neutralization reaction:

1) Word equation:

sulfuric acid + potassium hydroxide → potassium sulfate + water

↑ ↑ ↑ ↑

acid base salt water

2) Skeleton equation (unbalanced)

H₂SO₄ + KOH → K₂SO₄ + H₂O

#) Balanced chemical equation (including phases)

H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l) ← answer

Question 2:

1) Mol ratio:

Using the stoichiometric coefficients of the balanced chemical equation you get the mol ratio:

1 mol H₂SO₄ (aq) : 2 mol KOH (aq) : 1 mol K₂SO₄ (aq) : mol 2H₂O (l)

2) Moles of H₂SO₄:

V = 0.750 liter
M = 0.480 mol/liter
M = n/V ⇒ n = M×V = 0.480 mol/liter × 0.750 liter = 0.360 mol

3) Moles of KOH:

V = 0.700 liter
M = 0.290 mol/liter
M = n/V ⇒ n = M × V = 0.290 mol/liter × 0.700 liter = 0.203 mol

4) Determine the limiting reagent:

a) Stoichiometric ratio:

1 mol H₂SO₄ / 2 mol NaOH = 0.500 mol H₂SO4 / mol NaOH

b) Actual ratio:

0.360 mol H₂SO4 / 0.203 mol NaOH = 1.77 mol H₂SO₄ / mol NaOH

Since hte actual ratio of H₂SO₄ is greater than the stoichiometric ratio, you conclude that H₂SO₄ is in excess.

5) Amount of H₂SO₄ that reacts:

Since, KOH is the limiting reactant, using 0.203 mol KOH and the stoichiometryc ratio 1 mol H₂SO₄ / 2 mol KOH, you get:

x / 0.203 mol KOH = 1 mol H₂SO₄ / 2 mol KOH ⇒

x = 0.203 / 2 = 0.0677 mol of H₂SO₄

6) Concentration of H₂SO₄ remaining:

Initial amount - amount that reacted = 0.360 mol - 0.0677 mol = 0.292 mol

Total volume = 0.700 liter + 0.750 liter = 1.450 liter

Concetration = M

M = n / V = 0.292 mol / 1.450 liter = 0.201 M ← answer

6 0

3 years ago

In general how are word equations written to describe chemical equations

Alexus [3.1K]

chemicals reacting are written on the left, what is formed is written on the right after the = sign

eg Copper + oxygen = copper oxide.
Hope this helps , mark as brainliest and say thanks THANKS..

5 0

3 years ago

H-E-L-P!!...!! ʕ ꈍᴥꈍʔ

N76 [4]

Answer:

the correct answer is (2)

Explanation:

4 0

2 years ago

Read 2 more answers

The CO2 that builds up in the air of a submerged submarine can be removed by reacting it with sodium peroxide. 2 Na2O2 (s) + 2 C

grandymaker [24]

Answer: The sodium peroxide needed per sailor in a 24 hr period is 674 grams.

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 1.02 atm

V = Volume of gas = 150.0 ml = 0.15 L

n = number of moles of carbon dioxide = ?

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $20^0C=(20+273)K=293K$

$n=\frac{PV}{RT}$

$n=\frac{1.02atm\times 0.15L}{0.0821 L atm/K mol\times 293K}=0.006moles$

$2Na_2O_2(s)+2CO_2(g)\rightarrow 2Na_2CO_3(s)+O_2(g)$

According to stoichiometry:

2 moles of carbon dioxide require = 2 moles of sodium peroxide

Thus 0.006 moles carbon dioxide require= $\frac{2}{2}\times 0.006=0.006$ moles of sodium peroxide

Mass of sodium peroxide needed for 24 hour= $moles\times {\text {Molar mass}}=0.006mol\times 78g/mol\times 24\times 60=674g$

Thus sodium peroxide needed per sailor in a 24 hr period is 674 grams.

4 0

2 years ago