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Furkat [3]
3 years ago
7

When the reaction shown is correctly balanced, the coefficients are: kclo3 → kcl + o2?

Chemistry
2 answers:
REY [17]3 years ago
8 0
The balanced equation above will be;
2 KClO3 = 2 KCl +  3 O2 
Therefore, the coefficients are; 2,2,3
Which means 2 KClO3 will create 2 KCl and 3 oxygen molecules. 
notka56 [123]3 years ago
5 0
From the balanced equation 2KClO3 → 2KCl + 3O2, the coefficients are the following:
coefficient 2 in front of potassium chlorate KClO3
coefficient 2 in front of potassium chloride KCl 
coefficient 3 in front of oxygen molecule O2

We got this balanced equation by identifying the number of atoms of each element that we have in the given equation KClO3 → KCl + O2.
Looking at the subscripts of each atom on the reactant side and on the product side, we have
     KClO3 → KCl + O2
       K=1          K=1
       Cl=1         Cl=1
       O=3          O=2

We can see that the oxygens are not balanced. We add a coefficient 2 to the 3 oxygen atoms on the left side and another coefficient 3 to the 2 oxygen 
atoms on the right side to balance the oxygens:
     2KClO3 → KCl + 3O2
The coefficient 2 in front of potassium chlorate KClO3 multiplied by the subscript 3 of the oxygen atoms on the left side indicates 6 oxygen atoms just as the coefficient 3 multiplied by the subscript 2 on the right side indicates 6 oxygen atoms.

The number of potassium K atoms and chloride Cl atoms have changed as well:
     2KClO3 → KCl + 3O2
       K=2            K=1
       Cl=2          Cl=1
       O=6           O=6

We now have two potassium K atoms and two chloride Cl atoms on the reactant side, so we add a coefficient 2 to the potassium chloride KCl on the product side: 
     2KClO3 → 2KCl + 3O2, which is our final balanced equation.
        K=2           K=2
        Cl=2          Cl=2
        O=6           O=6
The potassium, chlorine, and oxygen atoms are now balanced.

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The volume of a gas at 7.00°c is 49.0 ml. if the volume increases to 74.0 ml and the pressure is constant, what will the tempera
marshall27 [118]
Using charles law
v1/t1=v2/t2
v1=49ml
v2=74
t1=7+273=280k
t2=?
49/280=74/t2
0.175=74/t2  cross multiply
0.175t2=74
t2=74/0.175
t2=422k or 149celcius
8 0
3 years ago
How many of these transition metals have you heard of before?
AleksAgata [21]

Answer:

38 elements or Element 31: Gallium Element 32: Germanium Element 33: Arsenic Element 34: Selenium Element 35: Bromine Element .

Explanation:

How many transition metals are there?

The 38 elements in groups 3 through 12 of the periodic table are called "transition metals". As with all metals, the transition elements are both ductile and malleable, and conduct electricity and heat.

6 0
3 years ago
Speed of an electron to orbit radius wire linear charge density is _________
ch4aika [34]

Answer:

A long uniformly charged wire has charge density λ=0.16μλ=0.16μC/m. 

7 0
3 years ago
A sample of a substance containing only magnesium and chlorine was tested in the lab and was found to be composed of 74.5% chlor
chubhunter [2.5K]
Use a ratio to find out that x/190.2 = 74.5/100 

hope this helps
8 0
3 years ago
A substance is analyzed and found to contain 85.7% carbon and 14.3% hydrogen by weight. A gaseous sample of the substance is fou
atroni [7]

Answer:

The empirical formula of the compound is CH2

Explanation:

<u>Step 1:</u> Data given

A substance contains 85.7 % carbon and 14.3 % hydrogen.

The substance has a density of 1.87 g/L

1 mol occupies 22.4 L

Molar mass of carbon = 12 g/mol

Molar mass of hydrogen = 1.01 g/mol

<u>Step 2</u>: Calculate molar mass of the substance

Since 1 mol occupies 22.4 L;

1 mol of this substance = 1.87g/L *22.4 = 41.888 grams

This means the molar mass of the substance is 41.888 g/mol

<u>Step 3:</u> Calculate mass of carbon:

85.8 % is carbon

this means 41.888 * 0.858 = 35.94 grams

<u>Step 4: </u>Calculate moles of carbon

moles C = mass C/ Molar mass C

Moles C = 35.94 grams / 12 g/mol

Moles C = 2.995 moles

<u>Step 5:</u> Calculate mass of hydrogen:

14.3 % is hydrogen

this means 41.888 * 0.143 = 5.99 grams

<u>Step 6 :</u>Calculate moles of hydrogen

Moles H  = 5.99 grams / 1.01 g/mol

Moles H = 5.93 moles

<u>Step 7: </u>Calculate  mol ratio

Ratio C:H = 1:2

The empirical formule = CH2

<u>Step 8</u>: calculate molar formule

Molar mass of empirical formule = 14.02 g/mol

n = Molar mass of substance / molar mass of empirical formule

n = 41.888 / 14.02 = 3

This means we have to multiply the empirical formula by 3

3*(CH2) = C3H6

C3H6 can be propene or cyclopropane

5 0
3 years ago
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