Answer:
Group 1 metals and transition metals are different from each other, mainly based on the colour of the chemical compounds that they form. The key difference between group 1 metals and transition metals is that the group 1 metals form colourless compounds, whereas the transition metals form colourful compounds.
CH₄(g) + 3 Cl₂(g) → CHCl₃(g) + 3 HCl(g)
From the equation we notice that 1 mole of methane produces 1 mole of chloroform:
16 g Methane → 119.38 g Chloroform
? g Methane → 37.5 g Chloroform
by cross multiplication:
= (16 * 37.5) / 119.38 = 5.0 g methane
Answer:
400 mL
Explanation:
Given data:
Mass of barium = 2.17 g
Pressure = 748 mmHg (748/760 = 0.98 atm)
Temperature = 21 °C ( 273+ 21 = 294k)
Milliliters of H₂ evolved = ?
Solution:
chemical equation:
Ba + 2H₂O → Ba(OH)₂ + H₂
Number of moles of barium:
Number of moles = mass/ molar mass
Number of moles = 2.17 g / 137.327 g/mol
Number of moles = 0.016 mol
Now we will compare the moles of barium with H₂.
Ba : H₂
1 : 1
0.016 : 0.016
Milliliters of H₂:
PV = nRT
V = nRT/P
V = 0.016 mol × 0.0821 atm. mol⁻¹.k⁻¹.L×294 k/0.98 atm
V = 0.39 atm. L/0.98 atm
V = 0.4 L
L to mL
0.4 × 1000 = 400 mL
Answer:
This question is incomplete.
Explanation:
This question is incomplete because of the absence of given mass and volume, however, the steps below will help solve the completed question. The molarity (M) of a solution is the number of moles of solute per liter of solvent. The formula is illustrated below;
Molarity = number of moles (n) / volume (in liter or dm³)
To calculate the number of moles of NaC₂H₃O₂, we say
number of moles (n) =
given or measured mass of NaC₂H₃O₂ ÷ molar mass of NaC₂H₃O₂
The volume of the solvent must be in liter (same as dm³). Thus, to convert mL to liter, we divide by 1000
The unit for Molarity is M (Molar concentration), mol/L or mol/dm³
Answer:
12 moles of CO₂.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
CO₂ + H₂O —> H₂CO₃
From the balanced equation above,
1 mole of CO₂ dissolves in water to produce 1 mole of H₂CO₃.
Finally, we shall determine the number of moles of CO₂ that will dissolve in water to produce 12 moles of H₂CO₃. This can be obtained as follow:
From the balanced equation above,
1 mole of CO₂ dissolves in water to produce 1 mole of H₂CO₃.
Therefore, 12 moles of CO₂ will also dissolve in water to produce 12 moles of H₂CO₃.
Thus, 12 moles of CO₂ is required.
