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jasenka [17]
3 years ago
11

What are the coefficients for the reaction _Cl2O5 + _H2O → _HClO3 once it is balanced?

Chemistry
2 answers:
netineya [11]3 years ago
6 0

Answer:

Answer is B. 1, 1, 2

Explanation:

<3

muminat3 years ago
5 0

Answer:

1, 1, 2

Explanation:

You might be interested in
How many moles of LiOH are needed to react completely with 25.5 g of CO2
LekaFEV [45]

Answer:

3.18 mol

Explanation:

2LiOH+CO_{2}-> Li_{2}CO_{3} +H_{2}O

n(CO2) = mass/ Mr.

             = 25.5 / 16

             = 1.59 mol

As per the equation above,

n(LiOH) : n(CO2)

     2      :    1

∴  3.18   :  1.59

     

3 0
3 years ago
Select all of the items that are true about a sample of water vapor at 101°C as it cools.
Dvinal [7]
<h3><u>Answer;</u></h3>

A) Its temperature will fall continuously until it condensed into a liquid.

<h3><u>Explanation</u>;</h3>
  • <em><u>Steam or water vapor is the gaseous state of liquid water.  When water vapor above a temperature of 100 degrees Celsius is cooled, the temperature falls continuously, and it undergoes condensation at a temperature of 100 degrees Celsius and turns into liquid water.</u></em>
  • The change of state from gaseous to liquid state occurs as a result of latent heat of vaporization that the water vapor carries.
4 0
3 years ago
Read 2 more answers
Who created the onion model in chemistry?
11111nata11111 [884]
Rick Maurer i think that’s how you spell his last name
8 0
3 years ago
If the energy of the reactants is less than
expeople1 [14]
The opposite type of reaction (where energy is taken in from the surroundings of a reaction and thus the energy of the reactants is lower than that of the products) is called an endothermic reaction
7 0
2 years ago
Calculate the pH of the solution that results when 20 mL of 0.2M HCOOH is mixed with 25mL of 0.2M NaOH solution.(post-equivalenc
Paul [167]

The pH of the solution : 12

<h3>Further explanation</h3>

Reaction

HCOOH    +     NaOH   ⇒     HCOONa   +   H₂O

mol HCOOH =

\tt 20~ml\times 0.2~M=4~mlmol

mol NaOH =

\tt 25~ml\times 0.2~M=5~mlmol

Mol NaOH>mol HCOOH ⇒ at the end of the reaction there will be a strong base remains from mol NaOH, so that the pH is determined from [OH⁻]

ICE method :

HCOOH    +     NaOH   ⇒     HCOONa   +   H₂O

4                          5

4                          4                     4                   4

0                          1                      1                    1

Concentration of [OH⁻] from NaOH :

\tt \dfrac{1~mlmol}{20+25~ml}=0.02

pOH=-log[OH⁻]

pOH=-log 10⁻²=2

pH+pOH=14

pH=14-2=12

3 0
3 years ago
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