Explanation:
A ____Chemical Reaction_______________________ is a well defined example of a chemical change. A chemical ___ _____chemical equation___________________ can be used to show the changes that occur in a chemical reaction. In a chemical reaction, the substance on the left side of the arrow are the starting substance. These substances are called ___Reactants________________________. The substances on the right side of the arrow are the substances that result from the reaction. These substances are called ____________Products_______________. The arrow is read as either produces or ______yields_____________________. According to the law of conservation of __________mass_________________, atoms are neither lost nor gained during a chemical reaction. This law is illustrated when a chemical equation is ________Balanced___________. When this is done, there will be the same number of ___________atoms________________ of each kind on both sides of the equation. In a chemical equation, the numbers that are placed in front of the symbols and the formulas are called ______________coefficients_____________. They are necessary to keep the ___________________________ of atoms in balance. There are several rules for balancing an equation. First, write the correct ____________(not so sure)_____________ for each reactant and product. Next, choose the coefficients that make the number of atoms of each _______elements(not so sure)________________ on each side of the equation equal. The correctly written formula should not be changed. If you change the formula of a substance, the equation is no longer ___________correct_____________. Changing a formula will indicate a ________Substance___________________ different than the one intended. To balance the equation Mg + O2 à MgO, first choose coefficients to make the number of atoms of each element on each side of the equation equal. You would need to place a coefficient of _________two___________
Answer:
Explanation:
(12 x 104 ) x (5 x 10-²) = 6 x 10 ³ 6 x 105 6.0x10²
1. (12 x 104 ) = 1248 or 1.248 x 10^3
2. (1.248 x 10^3)(5 x 10^-2) = 6.240 x 10^1 or 60 rounded to 1 sig fig
I don't understand "= 6 x 10 ³ 6 x 105 6.0x10² "
Answer:
54 days
Explanation:
We have to use the formula;
0.693/t1/2 =2.303/t log Ao/A
Where;
t1/2= half-life of phosphorus-32= 14.3 days
t= time taken for the activity to fall to 7.34% of its original value
Ao=initial activity of phosphorus-32
A= activity of phosphorus-32 after a time t
Note that;
A=0.0734Ao (the activity of the sample decreased to 7.34% of the activity of the original sample)
Substituting values;
0.693/14.3 = 2.303/t log Ao/0.0734Ao
0.693/14.3 = 2.303/t log 1/0.0734
0.693/14.3 = 2.6/t
0.048=2.6/t
t= 2.6/0.048
t= 54 days
