Answer:
Mass of aluminium in sample = 3.591 g ≅ 3.6 grams
Explanation:
Given that, A sample of aluminum absorbs 50.1 J of heat, upon which the temperature of the sample increases from 20.0°C to 35.5°C.
the specific heat of aluminum is 0.900 J/g- °C
The relation between heat absorbed and change in temperature is given by, Q = msΔT.
where Q = heat absorbed
m = mass of the substance
s = specific heat of substance
ΔT = change in temperature
Now, in our case, Q = 50.1 J ; s = 0.900 J/g- °C; ΔT= 35.5-20 = 15.5°C
⇒ m = 
⇒ m = 
= 3.591 g ≅ 3.6 g
⇒ m ≅ 3.6 g
Answer:
2311.2 cal
Explanation:
540 cal / g * 4.28 g = 2311.2 cal
Answer:
6.21 moles O
Explanation:
To find the moles of oxygen, you need to (1) convert grams Mn(ClO₄)₄ to moles Mn(ClO₄)₄ (via molar mass) and then (2) convert moles Mn(ClO₄)₄ to moles O (via mole-to-mole ratio from formula subscripts). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units.
Molar Mass (Mn(ClO₄)₄): 452.74 g/mol
1 Mn(ClO₄)₄ = 1 Mn and 4 Cl and 16 O
175.7 g Mn(ClO₄)₄ 1 mole 16 moles O
--------------------------- x ------------------- x --------------------------- = 6.21 moles O
452.74 g 1 mole Mn(ClO₄)₄
Answer:
160.9 mol ≅ 161.0 mol.
Explanation:
- It is known that every 1.0 mole of compound or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms (formula units).
Using cross multiplication:
1.0 mole of Fe(NO₃)₃ contains → 6.022 x 10²³ formula units.
??? mole of Fe(NO₃)₃ contains → 9.69 x 10²⁵ formula units.
<em>∴ The no. of moles of He contains (9.69 x 10²⁵ formula units)</em> = (1.0 mol)(9.69 x 10²⁵ formula units.)/(6.022 x 10²³ formula units) = <em>160.9 mol ≅ 161.0 mol.</em>
