Answer:
4. 15
Explanation:
The given formula is:- 
1 mole of the salt contains 2 moles of nitrogen atom, 8 moles of hydrogen, 1 mole of sulfur and 4 moles of oxygen atom as can be seen from the formula.
Thus,
1 mole of salt contains total of 15 moles of the atoms which constitute the salt.
<u>Hence, 4. is the answer.</u>
The red colour is the limiting reactant.
Red-blue colour ball and two white balls attached together are reactants.
Red-blue colour ball and two white and one red colour ball attached to each other are products.
<h3>What is a limiting reagent?</h3>
The reactant that is entirely used up in a reaction is called a limiting reagent.
A reactant is a substance that is present at the start of a chemical reaction. The substance(s) to the right of the arrow are called products.
A product is a substance that is present at the end of a chemical reaction.
Hence,
The red colour is the limiting reactant.
Red-blue colour ball and two white balls attached together are reactants.
Red-blue colour ball and two white and one red colour ball attached to each other are products.
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Answer:
2.99 M
Explanation:
In order to solve this problem we need to keep in mind the definition of molarity:
- Molarity = moles of solute / liters of solution
In order to calculate the moles of solute, we <u>convert 125.6 g of NaF into moles</u> using its <em>molar mass</em>:
- 125.6 g NaF ÷ 42 g/mol = 2.99 mol NaF
As the volume is already given, we can proceed to <em>calculate the molarity</em>:
- Molarity = 2.99 mol / 1.00 L = 2.99 M
Saturated hydrocarbons consists of C-C single bond whereas Unsaturated hydrocarbons consists C-C double/triple bond.
The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
The bond energies data is given as follows:
BE for C≡O = 1072 kJ/mol
BE for Cl-Cl = 242 kJ/mol
BE for C-Cl = 328 kJ/mol
BE for C=O = 766 kJ/mol
The enthalpy change for the reaction is given as :
ΔHr×n = ∑H reactant bond - ∑H product bond
ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )
ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )
ΔHr×n = 1314 - 1422
ΔHr×n = - 108 kJ
Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
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