Answer:
160.32 grams of Ca or 160 if rounded
Explanation:
Multiply moles of Ca by the conversion factor (molar mass of calcium) 40.08 g Ca/ 1 mol Ca, which then allows the cancelation of moles, leaving grams of Ca.
4 mol*40.08g/mol = 160.32 grams of Ca
please give thanks by clicking heart button :)
Answer:
D) N2O5
Explanation:
The molar mass of a substance is defined as the mass of this substance in 1 mol. To solve this question we must find the molar mass of each option:
<em>Molar mass NO:</em>
1N = 14g/mol*1
1O = 16g/mol*1
14+16 = 30g/mol
<em>Molar mass NO2:</em>
1N = 14g/mol*1
2O = 16g/mol*2
14+32 = 46g/mol
<em>Molar mass N2O:</em>
2N = 14g/mol*2
1O = 16g/mol*1
28+16 = 44g/mol
<em>Molar mass N2O5:</em>
2N = 14g/mol*2
5O = 16g/mol*5
28+80 = 108g/mol
That means the compound with the greatest mass is:
<h3>D) N2O5</h3>
Answer:
(See explanation for further details)
Explanation:
1) The quantity of moles of sulfur is:
2) The number of atoms of helium is:
3) The quantity of moles of carbon monoxide is:
4) The number of molecules of sulfur dioxide is:
5) The quantity of moles of sodium chloride is:
6) The number of formula units of magnesium iodide is:
7) The quantity of moles of potassium permanganate is:
8) The number of molecules of carbon tetrachloride is:
9) The quantity of moles of aluminium is:
10) The number of molecules of oxygen difluoride is:
<em><u>Question</u></em>
<em><u>What </u></em><em><u>does </u></em><em><u>it </u></em><em><u>mean </u></em><em><u>to </u></em><em><u>optimize</u></em><em><u> </u></em><em><u>a </u></em><em><u>solution?</u></em>
<em><u>To find out best possible solution for a given problem within the given constraint is generally termed as optimization</u></em>
<em><u>How </u></em><em><u>are </u></em><em><u>solution</u></em><em><u> </u></em><em><u>optimize</u></em><em><u> </u></em><em><u>?</u></em>
<em><u>To solve an optimization problem, begin by drawing a picture and introducing variables. Find an equation relating the variables. Find a function of one variable to describe the quantity that is to be minimized or maximized. Look for critical points to locate local extrema.</u></em>