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Mademuasel [1]
3 years ago
5

Evidence of chemical reactions include changes in

Chemistry
1 answer:
trasher [3.6K]3 years ago
7 0

Answer:

color, state, odor

Explanation:

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Compare and contrast the atomic structure of the chlorine-35 and chlorine-37 isotopes
katrin [286]

Chlorine-35 and 37 both have the same number of protons. Chlorine-37 has two more neutrons.

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3 years ago
Is concrete a homogeneous or heterogeneous mixture?
barxatty [35]
Heterogeneous mixture
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According to the MyPyramid plan, which foods should be the smallest portion of a teen’s diet?
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The answer is GRAINS 


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3 years ago
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Hello! Can you please help me do this question? I have to pick two but I’m not really sure what to pick..
In-s [12.5K]

Answer:

B and D

Explanation:

it is B and D because in no gravity  if you change the size of the ball it does not change anything. and if you swing fast it does not change anything but the speed. if you change the angle it is just the same modle but different angle

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3 years ago
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15.89 percent carbon, 21.18 percent oxygen, and 62.93 percent osmium. what is the empirical formula
otez555 [7]

Answer:

OsCO or COOs

Explanation:

Data given

Carbon = 15.89 %

Oxygen = 21.18 %

Osmium = 62.93%

Empirical formula = ?

Solution:

First find the masses of each component

Consider total compound is 100g

As we now

mass of element = % of component

So,

15.89 g of C     = 15.89 % Carbon

21.18 g of O      =   21.18 % Oxygen

62.93 g of Os  =   62.93% Osmium

Now convert the masses to moles

For Carbon

Molar mass of C = 12 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  15.89 g/ 12 g/mol

                  no. of mole =  1.3242

mole of C = 1.3242

For Oxygen

Molar mass of O = 16 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  21.18 g/ 16 g/mol

                  no. of mole =  

mole of O = 1.3238

For Os

Molar mass of Os = 190 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  62.93 g/ 190 g/mol

                  no. of mole =  

mole of Os = 1.3312

Now we have values in moles as below

C = 1.3242

O = 1.3238

Os = 1.3312

Divide the all values on the smalest values to get whole number ratio

C = 1.3242 /1.3238 = 1.0003

O = 1.3238 /1.3238 = 1

Os = 1.3312 /1.3238 = 1.0056

So all have round value 1 mole

C = 1

O = 1

Os = 1

So the empirical formula will be (OsCO) i.e. all 3 atoms in simplest small ratio

7 0
3 years ago
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