There are 2 significant figures.
"Rule 3: Trailing zeros are significant if a decimal point is shown in the number, but may or may not be significant if no decimal point is shown."
Answer:
7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100 M solution.
Explanation:
First of all the molecular weight of Na2SO4 is 142.08 gram.Now we all know that if the molecular weight of a compound is dissolved in 1000ml or 1 litee of water then the strength of that solution becomes 1 M.
According to the given question we have to prepare 0.100 M solution
1000 ml of solution contain 142.08×0.1= 14.208 gram Na2SO4
1 ml of solution contain 14.208÷1000= 0.014 gram
0.5L or 500ml of solution contain 0.014×500= 7gram Na2SO4.
So it can be stated that 7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100M solution.
Answer:
138 mg
Explanation:
A company is testing drinking water and wants to ensure that Ca content is below 155 ppm (= 155 mg/kg), that is, <em>155 milligrams of calcium per kilogram of drinking water</em>. We need to find the maximum amount of calcium in 890 g of drinking water.
Step 1: Convert the mass of drinking water to kilograms.
We will use the relation 1 kg = 1000 g.
Step 2: Calculate the maximum amount of calcium in 0.890 kg of drinking water
Gases, unlike solids and liquids, have neither fixed volume nor shape. They are molded entirely by the container in which they are held.
Answer : The amount of heat needed is, 1188 J
Explanation :
Formula used :
where,
q = heat needed = ?
m = mass of copper = 55 g
c = specific heat capacity of copper = 
= initial temperature = 
= final temperature = 
Now put all the given values in the above formula, we get:
Thus, the amount of heat needed is, 1188 J
