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>
O1Fl2
1. Assume an 100g sample, so the percentage will stay the same
2. Covert each element into their molar mass
29.6/16.00=1.8 mols of O
70.4/19.00=3.7 mols of Fl
3. Divide both by the smallest value of mol
1.8/1.8=1 O
3.7/1.8=2 Fl
4. Write the empirical formula:
O1Fl2
Answer:
M = 35 g/mol
Explanation:
Given data:
Density = 2.5 g/L
Pressure = 1.8 atm
Temperature = 30°C (273.15 +30 = 303.15 K)
Molar mass = ?
Solution:
d = PM/RT
M = dRT/P
M = 2.5 g/L × 0.0821 atm.L/mol.K × 303.15 K / 1.8 atm
M = 62.22 atm. g/mol/ 1.8 atm
M = 35 g/mol
Mixtures are classified as homogeneous and heterogeneous according to the distribution of the components of the mixture. If the components are uniformly distributed in the mixture resulting to uniform properties, then it is a homogeneous mixture. However, if it is not uniformly distributed where the composition of one part of the mixture is not the same as other parts, then it is classified as an homogeneous mixture.
