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anyanavicka [17]
3 years ago
6

How can the rate of a reaction be increased?

Chemistry
1 answer:
Nesterboy [21]3 years ago
6 0

Explanation:

Rate of reaction can be increased by having more surface area.

Therefore,

Option B is correct✔

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mestny [16]
It would be at 1 i believe because the heat you add after 0 degrees is then absorbed as potential energy instead of kinetic
3 0
3 years ago
How many molecules are there in 5H20?
Svetach [21]

Answer:

The number before any molecular formula applies to the entire formula. So here you have five molecules of water with two hydrogen atoms and one oxygen atom per molecule. Thus you have ten hydrogen atoms and five oxygen atoms in total.

3 0
3 years ago
Read 2 more answers
What volume is occupied by 2.719 x 1013 moles of methane gas, CH4?
slamgirl [31]

Answer:

6.5256×10¹⁴

Explanation:

n= v/24

v= n×25

v= 2.719×10¹³ × 24= 6.5256×10¹⁴ moles/dm³

hope it helps! please mark me brainliest

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7 0
3 years ago
Are fractions possible in oxidation state​
Andrew [12]

Answer:

No, have you ever listen that this compound has 1.5 electrons in outer most shell??? No, so oxidation state is always in integral number like, 1,2 3 etc

Explanation:

Order of reaction may be in fraction, but not oxidation state

3 0
3 years ago
The molecular weight of a gas is ________ g/mol if 3.5 g of the gas occupies 2.1 l at stp
bija089 [108]
<span>Pre-1982 definition of STP: 37 g/mol Post-1982 definition of STP: 38 g/mol This problem is somewhat ambiguous because the definition of STP changed in 1982. Prior to 1982, the definition was 273.15 K at a pressure of 1 atmosphere (101325 Pascals). Since 1982, the definition is 273.15 K at a pressure of exactly 100000 Pascals). Because of those 2 different definitions, the volume of 1 mole of gas is either 22.414 Liters (pre 1982 definition), or 22.71098 liters (post 1982 definition). And finally, there's entirely too many text books out there that still use the 35 year obsolete definition. So let's solve this problem using both definitions and you need to pick the correct answer for the text book you're using. First, determine how many moles of gas you have. Just simply divide the volume you have by the molar volume. Pre-1982: 2.1 / 22.414 = 0.093691443 moles Post-1982: 2.1 / 22.71098 = 0.092466287 moles Now determine the molar mass. Simply divide the mass by the moles. So Pre-1982: 3.5 g / 0.093691443 moles = 37.35666667 g/mol Post-1982: 3.5 g / 0.092466287 moles = 37.85163333 g/mol Finally, round to 2 significant figures. So Pre-1982: 37 g/mol Post-1982: 38 g/mol</span>
5 0
4 years ago
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