Your question is incomplete. However, I found a similar problem fromanother website as shown in the attached picture.
To solve this problem, you must know that at STP, the volume for any gas is 22.4 L/mol. So,
Moles O₂: 156.8 mL * 1 L/1000 mL* 1 mol/22.4 L = 0.007 moles
Mass calcium: 0.007 mol O₂ * 2 mol Ca/1 mol O₂ * 40 g/mol Ca =
<em> 0.56 g Ca</em>
Explanation:
Flourine has atomic number of 9 and hence 9 electrons in its neutral state. The full electronic configuration is given as;
1s2 2s2 2p5
Carbon has atomic number of 6 and hence 6 electrons in it's neutral state. The noble gas notation as the following format;
[closest noble gas before the element] remaining electrons
The nearest noble gas to carbon is Helium, the noble gas notation is given as;
[He] 2s4
Answer:
protactinium-234,
Explanation:
This is what happens when thorium-234 releases a W- boson, which then decays to an electron and an electron antineutrino.
The density of the block will be "7.11 gm/cm³". A further explanation is provided below.
Given values are:
Density of iron,
Mass of block of metal,
Volume of block,
By using the formula,
→ 
then,
→ The density of block will be:
= 
= 
= 
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Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:

Moles of glucose = 
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution = 
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution = 
Volume of the solution taken = 
Molarity of the solution after dilution = 
Volume of the solution after dilution= 



Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L

Moles of glucose = 
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.