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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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2 years ago
When an irregularly shaped chunk of an unknown metal with a mass of 25.32 g was placed in a graduated cylinder containing 25.00
inn [45]

Answer: The density of the unknown metal is 7.86 g/ml.

Explanation:

Density is defined as the mass contained per unit volume.

Density=\frac{mass}{Volume}

Given : Mass of metal  = 25.32 g

Volume of metal = volume of water displaced = (28.22 - 25.00) ml = 3.22 ml

Putting in the values we get:

density=\frac{25.32g}{3.22ml}

Density=7.86g/ml

Thus the density of the unknown metal is 7.86 g/ml

7 0
3 years ago
A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o
julsineya [31]

Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

Number \, of \, moles, n  = \frac{Mass}{Molar \ mass} =  \frac{5.84}{58.44} = 0.09993 \ moles \ of \ NaCl

0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

3 0
3 years ago
Please help. It's for chemistry
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