Answer:
13. 2g
14. 5000mL
15. 104,000m
16. 160mm
17. 5600000mg
18. 10hs
19. 0.250km
20. 1daL
Explanation:
13. 1000milligram (mg) = 1gram (g)
Hence, 2000mg = 2000/1000
= 2g
14. 1 litre (L) = 1000millilitre (mL)
Hence, 5L = 5 × 1000
= 5000mL.
15. 1kilometre (km) = 1000metre (m)
Hence, 104km = 104 × 1000
= 104,000m
16. 1 centimetre (cm) = 10millimeters (mm)
Hence, 16cm = 16 × 10
= 160mm
17. 1kilogram (kg) = 1000000 milligram (mg)
Hence, 5.6kg = 5.6 × 1000000
= 5600000mg
18. 1 second (s) = 0.01 hectosecond (hs)
Hence, 1000s = 1000 × 0.01
= 10hs
19. 1000metre (m) = 1kilometre (km)
Hence, 250m = 250/1000
= 0.250km
20. 1 centiliter (cl) = 0.001 Decaliter (daL)
Hence, 1000cl = 1000 × 0.001
= 1daL
A substance made of two or more elements chemically combined in a set ratio or proportion is called a : compound
For example, water is an example of a compound that made by two elements, Hydrogen and oxygen
hope this helps
Answer:
One way is to add or remove a product or a reactant in a chemical reaction at equilibrium. When additional reactant is added, the equilibrium shifts to reduce this stress: it makes more product. When additional product is added, the equilibrium shifts to reactants to reduce the stress. :)
Explanation:I am not sure how to use data on here :(
Answer:
0.85 mole
Explanation:
Step 1:
The balanced equation for the reaction of CaCl2 to produce CaCO3. This is illustrated below:
When CaCl2 react with Na2CO3, CaCO3 is produced according to the balanced equation:
CaCl2 + Na2CO3 -> CaCO3 + 2NaCl
Step 2:
Conversion of 85g of CaCO3 to mole. This is illustrated below:
Molar Mass of CaCO3 = 40 + 12 + (16x3) = 40 + 12 + 48 = 100g/mol
Mass of CaCO3 = 85g
Moles of CaCO3 =?
Number of mole = Mass /Molar Mass
Mole of CaCO3 = 85/100
Mole of caco= 0.85 mole
Step 3:
Determination of the number of mole of CaCl2 needed to produce 85g (i.e 0. 85 mole) of CaCO3.
This is illustrated below :
From the balanced equation above,
1 mole of CaCl2 reacted to produced 1 mole of CaCO3.
Therefore, 0.85 mole of CaCl2 will also react to produce 0.85 mole of CaCO3.
From the calculations made above, 0.85 mole of CaCl2 is needed to produce 85g of CaCO3