Hey there!
Mass = 5.6 Kg
Volume =8.2 L
D = m / V
D = 5.6 / 8.2
D = 0.6829 Kg/L
hope this helps!
Answer:
6.2g of NaBr are produced
Explanation:
The reaction of HBr with NaOH occurs as follows:
HBr + NaOH → NaBr + H2O
<em>Where 1 mole of each reactant produce 1 mole of NaBr</em>
To solve this question we need to find the moles of each reactant using their molar mass. With moles we can find limiting reactant and the moles (And mass) of NaBr produced, as follows:
<em>Moles HBr -Molar mass: 80.9119g/mol)-</em>
4.9g * (1mol/80.9119g) = 0.0606 moles HBr
<em>Moles NaOH -Molar mass: 40g/mol-</em>
3.86g * (1mol/40g) = 0.0965 moles NaOH
As the reaction is 1:1 and the moles of HBr < Moles NaOH, the limiting reactant is HBr and moles of NaBr produced are 0.0606 moles.
The mass of NaBr (Molar mass: 102.894g/mol) is:
0.0606 moles * (102.894g/mol) =
<h3>6.2g of NaBr are produced</h3>
Answer:
Cu + HNO3 → Cu(NO3)2 + <u>H2O</u> + <u>NO2</u>
Cu + 4HNO3 → Cu(NO3)2 + <u>2H2O</u> + <u>2NO2 </u>(balanced equation)
Answer:
(a) The concentration of after 5.00 min is 0.9672 mol/L
(b) The fraction of decomposed after 5.00 min is 0.568
The problem can be solved by using first order integrated reaction.
Explanation:
(a)
Rate constant of reaction = K =
Initial concentration of = 2.24 mol/L
Assuming final concentration of to be x mol/L
time (t) = 5.00 min
The first order integrated equation is shown below
Concentration of decomposed = 0.9672 mol/L
(b)
Concentration of decomposed =
Fraction of decomposed =
Answer:1.34 ×10^-5
Explanation:
AgCl = (Ag+) (Cl-)
=(S)(S)
= S²
S stands for molar solubility
Ksp =S² = 1.80 × 10^ -10
S= √[1.80 × 10^ -10]
=1.34 × 10^-5