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frosja888 [35]
3 years ago
9

Which of the following can be calculated from the mass of the reactants used in a chemical reaction?

Chemistry
2 answers:
borishaifa [10]3 years ago
6 0

Answer:

Three of these are correct.

Explanation:

Amount of limiting reactant used in a reaction

Theoretical yield of products

Amount of excess reactant from a reaction

Readme [11.4K]3 years ago
5 0

Answer:

points.

Explanation:

yeah........

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How many liters of hydrogen gas are required to react with 3.5" liters of oxygen gas in the following reaction? 2H2(g)+O2(g) --&
Hatshy [7]

Answer:

7 L of H₂.

Explanation:

The balanced equation for the reaction is given below:

2H₂ + O₂ —> 2H₂O

From the balanced equation above,

1 L of O₂ required 2 L of H₂.

Finally, we shall determine the volume of H₂ required to react with 3.5 L of O₂. This can be obtained as follow:

From the balanced equation above,

1 L of O₂ required 2 L of H₂.

Therefore, 3.5 L of O₂ will require

= 3.5 × 2 = 7 L of H₂.

Thus, 7 L of H₂ is required to for the reaction.

5 0
3 years ago
What is the correct clases for this drugs <br> Promethazine<br><br> heelp me please
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Answer:

MedlinePlus Drugs Information

Explanation:

Promethazine is in a class of medication called phenothiazines

6 0
3 years ago
What volume of 15.9 M Nitric acid would be required to make 7.2 L of 6.00 M nitric acid? (hint: dilution problem) will mark brai
irakobra [83]

Answer:

d

Explanation:

8 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
4 years ago
Zn(s)+cu2+(aq)→zn2+(aq)+cu(s). part a under standard conditions, what is the maximum electrical work, in joules, that the cell c
Marizza181 [45]
The energy in Joules is calculated as:

E = VQ

At standard conditions, the standard potential or the voltage is equal to 0.76 V. Next, we have to determine Q. The formula is as follows:

Q = nF, where n is the number of moles electron, while F is the Faraday's constant (96,500 C/mol e)

From the given reaction, notice that Zn was converted to Zn²⁺, while Cu²⁺ to Cu. So, 1 mol Cu = 2 mol e.

Q = 51 g Cu (1 mol Cu/63.55 g)(2 mol e/1 mol Cu)(96,500 C/mol e)
Q = 154885.92 C

Thus,
E = (0.76 V)(154885.92 C)
<em>E = 117,713.3 J</em>
4 0
4 years ago
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