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Lostsunrise [7]
3 years ago
6

The molar mass of H2O is 18.01 g/mol. The molar mass of O2 is 32.00 g/mol. What mass of H2O, in grams, must react to produce 50.

00 g of O2?
Chemistry
1 answer:
snow_lady [41]3 years ago
5 0

112.5 g. The production of 50.00 g O2 requires 112.5 g H2O.

a) Write the partially balanced equation for the decomposition of water.

MM = 18.02 32.00

2H2O → O2 + …

Mass/g = 50.00


b) Calculate the <em>moles of O2
</em>

Moles of O2 = 50.00 g O2 × (1 mol O2/16.00 g O2) = 3.1250 mol O2


c) Calculate the <em>moles of water</em>

Moles of H2O = 3.1250 mol O2 × (2 mol H2O/1 mol O2)

= 6.2500 mol H2O


d) Calculate the mass of water

Mass of H2O = 6.2500 mol H2O × (18.02 g H2O/1 mol H2O)

= 112.5 g H2O

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Answer:

Back titration is a titration done in reverse; instead of titrating the original sample, a known excess of standard reagent is added to the solution, and the excess is titrated.

I hope it's helpful!

3 0
2 years ago
Can someone help me with this
sergejj [24]

Answer: 5. Positive

6. Negative

9. Oxygen carbon hydrogen nitrogen

Explanation:

7 0
2 years ago
|
lilavasa [31]

1.66 M is the concentration of the chemist's working solution.

<h3>What is molarity?</h3>

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

In this case, we have a solution of Zn(NO₃)₂.

The chemist wants to prepare a dilute solution of this reactant.

The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.

We want to know the concentration of this diluted solution.

As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:

n_1= n_2  (1)

and we also know that:

n = M x V_2

If we replace this expression in (1) we have:

M_1 x V_1= M_2 x V_2

Where 1, would be the stock solution and 2, the solution we want to prepare.

So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is M_2:

4.93 x 210 =  620 xM_2

M_2 = 1.66 M

This is the concentration of the solution prepared.

Learn more about molarity here:

brainly.com/question/19517011

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6 0
2 years ago
Comparing the 2-bromobutane + methoxide and 2-bromobutane + t-butoxide reactions, choose the statements that BEST describe the d
Elina [12.6K]

Answer:

an increase in 1-butene was observed when t-butoxide was used

Explanation:

When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.

Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.

The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.

The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.

8 0
3 years ago
Hydrates that have a low vapor pressure and remove moisture from air are said to be ___. Question 8 options: effloresce hygrosco
-Dominant- [34]

Answer:

Hygroscopic

Explanation:

An hygroscopic substance is one that absorbs moisture from the atmosphere and becomes wet. Their ability to remove water from air is less than that of deliquescent substances. Most of the solid hygroscopic substances forms pasty substances and not solutions like the deliquescent compounds.

Examples are sodium trioxonitrate(v), copper(ii) oxide e.t.c

Efflorescence compounds gives off their water of crystallization to the atmosphere.

4 0
3 years ago
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