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Eduardwww [97]
2 years ago
13

What is the percent composition of chlorine in the compound Ba(ClO3)2?

Chemistry
1 answer:
tia_tia [17]2 years ago
6 0

You're going to divide the mass of chlorine within the compound by the mass of the compound, and then multiply the result by 100 to get the answer

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The volume of a sample of gas is 0.5 L and the pressure is 0.98 atm. The volume is increased to 1.0 L. Show the set
meriva

Given,

P1 = 0.98 atm

V1 = 0.5 L

V2 = 1.0 L

P2 = ?

Solution,

According to Boyle's Law,

P1V1 = P2V2

0.98 × 0.5 = 1.0 × P2

P2 = 0.98 × 0.5 × 1.0

P2 = 0.49 atm

Answer - The new pressure is 0.49 atm.

5 0
3 years ago
Please answer this Q:
Anarel [89]
D. container four because of the water
3 0
2 years ago
Describe the day of June 21 at the North Pole in terms of daylight and darkness <br> ( science )
frosja888 [35]
On June 21, as seen from the North pole ...

-- the sun has been up, and it's been light outside,
for the past three months ... ever since March 21 . 

-- The sun won't set, and it won't be dark outside,
for another three months ... until September 21.

-- Here at the North pole, it stays daylight for six months straight.
Today, on June 21, we're exactly halfway through the period of
continuous daylight.
7 0
3 years ago
Look at the following enthalpy diagram. Select all that apply.
Drupady [299]

Answer:

Option 2 and 4 are correct

Explanation:

The reactants in the attached image have more enthalpy and hence less stability as they are more reactive. Thus, Product is more stable than the reactants.

This is an addition reaction in which two reactants add up to form the product.

Very less activation energy is required as the reactants themselves are unstable, possess high energy and hence are very reactive.

Reactants have more energy than the products.  

5 0
2 years ago
Calculate the standard emf for the following reaction:
krek1111 [17]
In order to solve this, we need to know the standard cell potentials of the half reaction from the given overall reaction.
The half reactions with their standard cell potentials are:
<span>2ClO−3(aq) + 12H+(aq) + 10e- = Cl2(g) + 6H2O(l) 
</span><span>E = +1.47 
</span>
<span>Br(l) + 2e- = 2Br- 
</span><span>E = +1.065
</span>
We solve for the standard emf by subtracting the standard emf of the oxidation from the reducation, so:
1.47 - 1.065 = 0.405 V
4 0
3 years ago
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