Answer:
802.69 g
Explanation:
The molar mass of Barium nitrite is 229.34 g/mol, so 3.5 moles of it will have a mass of ...
3.5 mol × 229.34 g/mol = 802.69 g
Almost all hydrocarbon 'burn' reactions involve oxygen; it's by far the most reactive substance in air.
<span>Hydrocarbon combustions always involve </span>
<span>[some hydrocarbon] + oxygen --> carbon dioxide + steam. </span>
C6H6(l) + O2 (g)--> CO2 (g)+ H2O (g)
<span>Balance carbon, six on each side: </span>
C6H6(l) + O2 (g)--> 6CO2 (g)+ H2O (g)
<span>Balance hydrogen, six on each side: </span>
C6H6(l) + O2 (g)--> 6CO2(g) + 3H2O (g)
<span>Now, we have fifteen oxygens on the right and O2 on the left. </span>
<span>Two ways to deal with that. We can use a fraction: </span>
C6H6 (l)+ (15/2)O2 (g)--> 6CO2 (g)+ 3H2O (g)
<span>Or, if you prefer to have whole number coefficients, double everything </span>
<span>to get rid of the fraction: </span>
2C6H6 (l)+ 15O2 (g)--> 12CO2 (g)+ 6H2O (g)
<span>With the SATP states thrown in... </span>
C6H6(l) + (15/2)O2(g) --> 6CO2(g) + 3H2O(g)
Answer:
14.3mL you require to reach the half-equivalence point
Explanation:
A strong acid as HClO₄ reacts with a weak base as CH₃CH₂NH₂, thus:
CH₃CH₂NH₂ + HClO₄ → CH₃CH₂NH₃⁺ + ClO₄⁻
As the reaction is 1:1, to reach the equivalence point you require to add the moles of HClO₄ equal to moles CH₃CH₂NH₂ you add originally. Also, half-equivalence point requires to add half-moles of CH₃CH₂NH₂ you add originally.
Initial moles of CH₃CH₂NH₂ are:
20.8mL = 0.0208L × (0.51mol CH₃CH₂NH₂ / 1L) =
0.0106moles CH₃CH₂NH₂
To reach the half-equivalence point you require:
0.0106moles ÷ 2 = 0.005304 moles HClO₄
As concentration of HClO₄ is 0.37M, volume you require to add 0.005304moles is:
0.005304 moles HClO₄ ₓ (1L / 0.37mol) = 0.0143L =
<h3> 14.3mL you require to reach the half-equivalence point</h3>
Answer:
354.67K
Explanation:
Applying
P₁V₁/T₁ = P₂V₂/T₂................. Equation 1
Where Where P₁ = initial pressure, T₁ = Initial temperature, V₁ = Initial Volume, P₂ = Final pressure, V₂ = Final Volume, T₂ = Final Temperature.
From the question, we are ask to look for the final temperature,
Therefore we make T₂ the subject of the equation
T₂ = P₂V₂T₁/P₁V₁............. Equation 2
Given: P₁ = 600 kPa, V₁ = 500 mL, T₁ = 77 °C = (273+77) = 350 K, P₂ = 760 kPa, V₂ = 400.0 mL
Substitute these values into equation 2
T₂ = (760×400×350)/(600×500)
T₂ = 354.67 K
Speed ; the rate at which someone or something is able to move or operate.
Motion; the action or process of moving or being moved
