44g of CO2 can produce by the reaction of carbon with oxygen
Answer:
14. 13.2cg = 1.32dg
15. 3.8m = 0.0038km
16. 24.8L = 24800mL
17. 0.87kL = 870L
18. 26.01cm = 0.0002601km
19. 0.001hm = 10cm
Explanation:
14. 13.2/10 = 1.32
15. 38/1000 = 0.0038
16. 24.8(1000) = 24,800
17. 0.87(1000) = 870
18. 26.01/100000 = 0.0002601
19. 0.001hm(10000) = 10
An easy way to do these by yourself is to familiarize yourself with what each prefix means. Once you do this, you can multiply the value of the prefix when converting from a smaller unit of measurement to a larger one and divide the value of the prefix when converting from a large unit of measurement to a smaller one.
Answer:
The correct answer is - 4.
Explanation:
As we known and also given that the total of the superscripts that is mass numbers, A in the reactants and products must be the same.The mass of products A can understand and calculated by this -
The sum of the product mass number of products = mass of reactant
237Np93 →233 Pa91 +AZX is the equation,
Solution:
Mass of reactants = 237
Mass of products are - Pa =233 and A = ?
233 + A = 237
A = 237 - 233
A = 4
So the equation will be:
237Np93 →233 Pa91 +4He2 (atomic number Z = 2 ∵ difference in the atomic number of reactant and products)
Percent yield is expressed as the ratio of the actual yield and the theoretical yield of the reaction multiplied by 100 to get the percent value. The actual yield is usually given in a problem. The theoretical yield is calculated from the reaction. For this problem, it cannot be solved since we cannot obtain the theoretical yield.
Answer: Option (b) is the correct answer.
Explanation:
The given chemical reaction shows that hydrogen cyanide acid has been added to water which results in the formation of hydronium ion and cyanide ion.
Also, when we add a base like sodium hydroxide (NaOH) to HCN then it will help in accepting a proton (
) from hydrogen cyanide. As a result, formation of 
anion will be rapid and easy.
This will make the system not to do any extra work. So, amount of work done by system will decrease.
Thus, we can conclude that out of the given options, add solid NaOH to the reaction (assume no volume change) will decrease the amount of work the system could perform.
