Answer:
You can find in the given attachemnet
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___________
<u>Answer:</u> The final temperature of the copper is 95°C.
<u>Explanation:</u>
To calculate the final temperature for the given amount of heat absorbed, we use the equation:

Q = heat absorbed = +133 J (heat is added to the system)
m = mass of copper = 5.00 g
c = specific heat capacity of copper = 0.38 J/g ° C


Putting values in above equation, we get:

Hence, the final temperature of the copper is 95°C.
Answer:
Alkali metals are soft and have low melting points.
Answer:
a)23.2 L
b)68.3kPa
c)7.5 atm
d)60.5L
e)1.67 atm
Explanation:
From Boyle's law:
P1V1=P2V2
P1= 748mmHg
P2=725mmHg
V1= 22.5L
V2??
V2= P1V1/P2= 748×22.5/725= 23.2 L
b)
V1=4.0L
P1= 205×10^3Pa
V2= 12.0L
P2=???
P2= P1V1/V2= 205×10^3×4/12
P2= 68.3×10^3 Pa or 68.3kPa
c)
P1= 1 atm
V1= 196.0L
P2= ??
V2= 26.0L
P2= P1V1/V2=1×196.0/26.0
P2= 7.5 atm
d)
V1= 40.0L
P1= 12.7×10^3Pa
V2=???
P2= 8.4×103Pa
V2= P1V1/P2= 12.7×10^3×40.0/8.4×103
V2=60.5L
e)
V1= 100mL
P1= 1atm
V2= 60mL
P2=???
P2= P1V1/V2= 1×100/60
P2= 1.67 atm