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Masteriza [31]
3 years ago
8

12.00 moles of NaClO3 will produce how many grams of O2?

Chemistry
1 answer:
abruzzese [7]3 years ago
5 0

A) 256g of O2, hope this helps :)

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What is the temperature of 0.500 moles of helium that occupies a volume of 15.0 L at a pressure of 1.6 atm?
denis23 [38]

Answer:

600K

Explanation:

PV=nRT

T=PV/nR

= 1.6atm* 15.0L/ 0.5mol*0.0821LatmK^-1mol^-1

=600K

6 0
3 years ago
Express the following numbers as decimals: (a) 1.52 x 10^-2, (b) 7.78 x 10^-8, (c) 1 x 10^-6, (d) 1.6001 x 10^3.
Alchen [17]

Answer :

(a) 0.0152

(b) 0.0000000778

(c) 0.000001

(d) 1600.1

Explanation :

Scientific notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

For example :

5000 is written as 5.0\times 10^3

889.9 is written as 8.899\times 10^{-2}

In this examples, 5000 and 889.9 are written in the standard notation and 5.0\times 10^3  and 8.899\times 10^{-2}  are written in the scientific notation.

(a) 1.52\times 10^{-2}

The standard notation is, 0.0152

(b) 7.78\times 10^{-8}

The standard notation is, 0.0000000778

(c) 1\times 10^{-6}

The standard notation is, 0.000001

(d) 1.6001\times 10^{3}

The standard notation is, 1600.1

5 0
3 years ago
How does a heat affect the chemical reaction? ​
patriot [66]

Answer:

Increasing the temperature increases reaction rates because of the disproportionately large increase in the number of high energy collisions. It is only these collisions (possessing at least the activation energy for the reaction) which result in a reaction.

Explanation:

6 0
3 years ago
Read 2 more answers
2) Calculate the mass(g) of the following substances:<br> b) 0.119 mole MgCl2
VARVARA [1.3K]

Answer:

MgCl2 = 24 + 2(35.5)

= 95

mass of substance = mol × molar mass

= 0.119 × 95

= 11.305 g

4 0
3 years ago
A standard 10.00 g mass is weighed on an analytical balance 100 times. The average and standard deviation obtained gives 10.12 ±
Mekhanik [1.2K]

Answer:

There was an improvement in accuracy. There was no change in precision.

Explanation:

<em>The average mass after recalibration is closer to the mass of the standard, </em>so the recalibration improved the accuracy<em> </em>(the measurement is closer to an accepted 'true' value).

The standard deviation did not change, so the precision (or how disperse the measurements are) was not affected.

5 0
3 years ago
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