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natali 33 [55]
3 years ago
11

How many moles of water can be formed from 0.57 moles of hydrogen gas?

Chemistry
1 answer:
BartSMP [9]3 years ago
7 0

Answer:

0.57 water

Explanation:

To solve this problem, we need to write the reaction expression first.

The reactants are oxygen gas and hydrogen gas.

They react to give a product of water

       2H₂    +    O₂   →   2 H₂O  

Given that;

Number of moles of hydrogen gas = 0.57moles

From the balanced reaction expression;

       2 moles of hydrogen gas produces 2 moles of water

   So;

    0.57mole of hydrogen gas will also produce 0.57 water

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Answer:

A supersaturated solution

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a 6.7 volume of air, initially at 23 degrees celsius and .98 atm, is compressed to 2.7 L while heated to 125 degrees celsius. Wh
Annette [7]
Data:

V1 = 6.7 liter
T1 = 23° = 23 + 273.15 K = 300.15 K
P1 = 0.98 atm

V2 = 2.7 liter
T2 = 125° = 125 + 273.15 K = 398.15 K
P2 = ?

Formula:

Combined law of ideal gases: P1 V1 / T1 = P2 V2 / T2

=> P2 = P1 V1 T2 / (T1 V2)

P2 = 0.98 atm * 6.7 liter * 398.15 K / (300.15K * 2.7 liter)

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3 years ago
A person's heartbeat is 68 beats per minute. If his/her heart beats 3.1e9 times in a lifetime, how long (in whole years) does th
kari74 [83]
To determine the number of years for a person to live, we can divide the total number of beats in a lifetime to the number of beats per minute. We first need to check if the units are similar so we can cancel them. We do as follows:

3.1x10^9 beats / 68 beats per minute = 45588235.29 minutes ( 1hr / 60 min ) ( 1 day/24hr) ( 1 year / 365 days ) = 87 years

Hope this answers the question. Have a nice  day.
8 0
3 years ago
Read 2 more answers
A solution contains 0.021 M Cl and 0.017 M I. A solution containing copper (I) ions is added to selectively precipitate one of t
lidiya [134]

<u>Answer:</u> Copper (I) iodide will precipitate first.

<u>Explanation:</u>

We are given:

K_{sp} of CuCl = 1.0\times 10^{-6}

K_{sp} of CuI = 5.1\times 10^{-12}

Concentration of Cl^-\text{ ion}=0.021M

Concentration of I^-\text{ ion}=0.017M

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

  • <u>For CuCl:</u>

K_{sp}=[Cu^+][Cl^-]

Putting values in above equation, we get:

1.0\times 10^{-6}=[Cu^+]\times 0.021

[Cu^+]=\frac{1.0\times 10^{-6}}{0.021}=4.76\times 10^{-5}M

Concentration of copper (I) ion = 4.76\times 10^{-5}M

  • <u>For CuI:</u>

K_{sp}=[Cu^+][I^-]

Putting values in above equation, we get:

5.1\times 10^{-12}=[Cu^+]\times 0.017

[Cu^+]=\frac{5.1\times 10^{-12}}{0.017}=3.00\times 10^{-10}M

Concentration of copper (I) ion = 3.00\times 10^{-10}M

For the precipitation of copper (I) ions, we need less concentration of copper (I) ions. So, copper (I) iodide will precipitate first.

7 0
3 years ago
For the rxn CaCo3(s)+2Hcl(aq)_CaCl(aq)+Ca2(g)+H2O(l)68.1 g solid CaCo3 is mixed with 51.6Hcl what number of grams of Co2 will be
Svetach [21]

Answer:

29.9 g of CO₂ will be produced by the reaction.

Explanation:

This is the reaction:

CaCO₃(s)  + 2HCl (aq)  →  CaCl₂(aq)  + CO₂(g) +  H₂O(l)

First of all, we state the moles of each reactant:

68.1 g . 1mol/ 100.08g = 0.680 mol of carbonate

51.6 g . 1 mol/36.45g = 1.46 mol of acid.

The solid salt is the limiting reactant. Ratio is 2:1

2 moles of acid can react to 1 mol of salt

1.46 mol of acid may react with (1.46 . 1)/2 = 0.727 moles

As we only have 0.680 moles of salt, we do not have enough.

Let's work at the product side. Ratio is 1:1

1 mo of salt can produce 1 mol of gas

0.680 moles will produce 0.680 moles of gas

We convert the moles to mass → 0.680 mol . 44g / 1mol = 29.9 g of CO₂

8 0
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