Answer:
Molarity = moles ÷ liters
to get moles of NaBr divide grams of NaBr by its molar mass (mass of Na + mass of Bromine)
Na = 22.989769
Br = 79.904
molar mass of NaBr = 102.893769
6.6g ÷ 102.893769 = 0.064143826 moles of NaBr
0.064143826 moles ÷ 0.60 liters = 0.1069 molar concentration or 11 %
Answer: 8556 mm, or 855.6 cm (8560 mm to 3 sig figs)
Explanation: Convert mm to cm by dividing by 10 (1cm/10mm)
Find the area of the foil face in cm^2 (30cm*0.2020cm) = 0.606 cm^2
Calculate the volume occupied by 1.40 kg of foil in cm^3. 1.40kg = 1400g
1.400g/(2.7 g/cm^3) = 518.5 cm^3 for 1.40 kg Au
Volume = Area (of the face) * Length
We want Length:
Length = Volume/Area
L = (518.5 cm^3/0.606 cm^2)
L = 855.6 cm (8556 mm) Round to 3 sig figs (856 cm and 8560 mm)
Answer:
K⁺ (aq) + F⁻ (aq) + H⁺ (aq) + Cl⁻ (aq) → KCl (aq) + H⁺ (aq) + F⁻ (aq)
Explanation:
KF (aq) + HCl (aq) → KCl (aq) + HF (aq)
KF (aq) → K⁺ (aq) + F⁻ (aq)
HCl (aq) → H⁺ (aq) + Cl⁻ (aq)
KCl (aq) → K⁺ (aq) + Cl⁻ (aq)
HF (aq) → H⁺ (aq) + F⁻ (aq)
Answer:
Flammability is a material’s ability to burn in the presence of <u><em>oxygen.</em></u>
Explanation:
Flammability can be described as the ability of a substance to get ignited. Flammability will lead to fire or combustion. Some substances are highly flammable like Benzene. Other tend to be just flammable. And there are also compounds which will nor be flammable at all as they won't react with oxygen. Examples of these substances include helium, steel or glass.
The flammability of a substance shall be considered a very important aspect when storing or transporting a substance.
Answer:
q = -6464.9 kJ
Explanation:
We are given that the heat of combustion is ∆H° = −394 kJ per mol of carbon.Therefore what we need to do is calculate how many moles of C are in the lump of coal by finding its mass since the density is given.
vol = 5.6 cm x 5.1 cm x 4.6 cm = 131.38 cm³
m = d x v = 1.5 g/cm³ x 131.38 cm³ = 197.06 g
mol C = m/MW = 197.06 g/ 12.01g/mol = 16.41 mol
q = −394 kJ /mol C x 16.41 mol C = -6464.9 kJ