Answer:
There is only one formula to use and we should assume ideal gas. This equation is: PV=nRT. For the following questions manipulate this equation to get the answer.
1. n = PV/RT = (249*1000 Pa)(15.6 L)(1 m^3/1000 L)/(8.314 Pa-m^3/mol-K))(21+273) = 1.59 mol
2. P = nRT/V = (1.59)(8.314)(51+273)/(15.6/1000)(1000) = 274.55 kPa
3. Since the answer in #2 is more than 269 kPa, then the tires will likely burst. 4. Reduce pressure way below the limit 269 kPa.
Explanation:
Answer:
The answer is C. Gas particles have no attractive forces between them.
Explanation:
Excess reactant : Na
NaCl produced : = 16.497 g
<h3>Further explanation</h3>
Given
Reaction(balanced)
2Na + Cl₂⇒ 2NaCl
20 g Na
10 g Cl₂
Required
Excess reactant
NaCl produced
Solution
mol Na(Ar = 23 g/mol) :
= 20 : 23 = 0.87
mol Cl₂(MW=71 g/mol):
= 10 : 71 g/mol = 0.141
mol : coefficient :
Na = 0.87 : 2 = 0.435
Cl₂ = 0.141 : 1 = 0.141
Limiting reactant : Cl₂(smaller ratio)
Excess reactant : Na
Mol NaCl based on mol Cl₂, so mol NaCl :
= 2/1 x mol Cl₂
= 2/1 x 0.141
= 0.282
Mass NaCl :
= 0.282 x 58.5 g/mol
= 16.497 g
Answer:
0.109 g.
Explanation:
Equation of the reaction:
Na3PO4 + 3HCl --> 3NaCl + H3PO4
Number of moles of HCl = molar concentration × volume
= 0.1 × 0.04
= 0.004 mol.
By stoichiometry, 1 mole of Na3PO4 neutralises 3 moles of HCl. Therefore, number of moles of Na3PO4 = 0.004/3
= 0.0013 mol
Mass of Na3PO4 = molar mass × number of moles
= 0.0013 × 164
= 0.219 g
Since 50% of Na3PO4 was present in the sample. Let 100 g be the total mass of the substance
= 0.219 × 50 g/100 g
= 0.109 g.
Answer:
<h3>The answer is 8.29 %</h3>
Explanation:
The percentage error of a certain measurement can be found by using the formula
From the question
actual density = 19.30g/L
error = 20.9 - 19.3 = 1.6
We have
We have the final answer as
<h3>8.29 %</h3>
Hope this helps you
