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Nimfa-mama [501]
3 years ago
14

2 NaOH + H2SO4 → 2 H,0 + Na S04

Chemistry
1 answer:
maw [93]3 years ago
6 0

Answer:

200 grams of sodium  hydroxide produce 595 grams of sodium sulfate if we have an excess of sulphuric acid

Explanation:

2 NaOH + H2SO4 → 2 H20 + Na S04

From the above chemical equation we can state that

2 moles of sodium  hydroxide reacts with 1 mole of  sulfuric acid to produce 1 mole of sodium sulfate.

Now if we  start with 200 grams of sodium  hydroxide and you have an excess of sulfuric acid then the amount in grams of sodium sulfate will be formed

2 moles of sodium  hydroxide produce 1 mole of sodium sulfate.

In term of mass, we can say that

40 grams of sodium  hydroxide produce 119grams of sodium sulfate

1 grams of sodium  hydroxide produce (119/40) grams of sodium sulfate

then 200 grams of sodium  hydroxide produce ((119*200)/40) grams of sodium sulfate

or 200 grams of sodium  hydroxide produce 595 grams of sodium sulfate

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Help plz I will give u BRAINLIST
Agata [3.3K]

CS2 + 3O2 = CO2 + 2SO2

1 mole of CS2 gives 1 mole of CO2

12 + 2(32) = 76g of CS2  yields 44 g of CO2

Theoretically 1 g of CS2 yields 44/76 g CO2

Therefore 50 g CS2 should yield    50*44 / 76 = 28.95 g

So % yield = 103.6 %  ( which is not possible  because you can't create matter from nothing).

The 30g cannot be right . This is experimental err.

6 0
3 years ago
The work function is the energy that must be supplied to cause the release of an electron from a photoelectric material. The cor
vladimir2022 [97]

Answer:

No photoelectric effect is observed for Mercury.

Explanation:

From E= hf

h= Plank's constant

f= frequency of incident light

Threshold Frequency of mercury= 435×10^3/ 6.6×10^-34 × 6.02×10^23

f= 11×10^14 Hz

The highest frequency of visible light is 7.5×10^14. This is clearly less than the threshold frequency of mercury hence no electron is emitted from the mercury surface

3 0
3 years ago
How to convert 2750 mg to g?
choli [55]

Answer:

2.75 g

Explanation:

divide 2750 by 1000

3 0
3 years ago
Perform the following operation
lukranit [14]

Answer:

(5.4 x 10³) x (1.2 x 10⁷) = 6.48 x 10¹⁰

With correct significant figures, the answer would be 6.5 x 10¹⁰.

4 0
2 years ago
8.0 mol AgNO3 reacts with 5.0 mol Zn in
Yakvenalex [24]

Taking into account the reaction stoichiometry, 8 moles of Ag can be produced from 8 moles of AgNO₃ and 5 moles of Zn.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 AgNO₃ + Zn → 2 Ag + Zn(NO₃)₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • AgNO₃: 2 moles
  • Zn: 1 mole
  • Ag: 2 moles
  • Zn(NO₃)₂: 1 mole

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Zn reacts with 2 moles of AgNO₃, 5 moles of Zn reacts with how many moles of AgNO₃?

amount of moles of AgNO_{3}= \frac{5 moles of Znx2 moles of AgNO_{3}}{1 mole of Zn}

<u><em>amount of moles of AgNO₃= 10 moles </em></u>

But 10 moles of AgNO₃ are not available, 8 moles are available. Since you have less moles than you need to react with 5 moles of Zn, AgNO₃ will be the limiting reagent.

<h3>Moles of Ag formed</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 2 moles of AgNO₃ form 2 moles of Ag, 8 moles of AgNO₃ form how many moles of Ag?

amount of moles of Ag=\frac{8 moles of AgNO_{3}x2 moles of Ag }{2 moles of AgNO_{3}}

<u><em>amount of moles of Ag= 8 moles</em></u>

Then, 8 moles of Ag can be produced from 8 moles of AgNO₃ and 5 moles of Zn.

Learn more about the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

#SPJ1

4 0
1 year ago
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