C3H8+3O2--->3CO2+8H
Therefore for every 1:3 there are 3 Carbon dioxides that form. That means find the limiting reactant from the two reactants.
5.5g(1mole C3H8/44.03g of C3H8)=0.1249 moled of C3H8 and if for every one C3H8 we can form three CO2. We can assume 0.3747 miles of CO2 will be produced.
15g of O2(1 mole O2/32g of O2)=0.4685moles O2 and if for every three O2 we can produce three CO2 we may assume a 1:1 ratio.
This means C3H8 will be your limiting reactant. Therefore 0.3747 moles of CO2 will be produced.
0.3747 moles of CO2(48.01 g of CO2/1 mole of CO2)= 17.99 grams of CO2
Answer:
Explanation:
<u>1. Word equation:</u>
- <em>mercury(II) oxide → mercury + oxygen </em>
<u>2. Balanced molecular equation:</u>
<u>3. Mole ratio</u>
Write the ratio of the coefficients of the substances that are object of the problem:
<u>4. Calculate the number of moles of O₂(g)</u>
Use the equation for ideal gases:
<u>5. Calculate the number of moles of HgO</u>
<u>6. Convert to mass</u>
- mass = # moles × molar mass
- molar mass of HgO: 216.591g/mol
- mass = 0.315mol × 216.591g/mol = 68.3g
Answer:
<h3>Hlo there !! </h3>
<u>One mole of any substance contains 6.022*1023 structural units (atoms, molecules, ions, etc.). This number is known as the Avogadro constant.</u>
<u>One mole of any substance contains 6.022*1023 structural units (atoms, molecules, ions, etc.). This number is known as the Avogadro constant.So 1.04*107 mol of Al contains 1.40*107 * 6.022*1023 = 8.43*1030 structural units (in case of Al – atoms).</u>
<h3><u>8.43*1030 particles Al.</u></h3>
Explanation:
<h3>Hope this helps !!</h3>
Option c
water is non polar and can dissolve non polar compounds such as oil
water is unable to dissolve or mix with substances like oil therefore making the statement false
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