0.208 is the specific heat capacity of the metal.
Explanation:
Given:
mass (m) = 63.5 grams 0R 0.0635 kg
Heat absorbed (q) = 355 Joules
Δ T (change in temperature) = 4.56 degrees or 273.15+4.56 = 268.59 K
cp (specific heat capacity) = ?
the formula used for heat absorbed and to calculate specific heat capacity of a substance will be calculated by using the equation:
q = mc Δ T
c = 
c = 
= 0.208 J/gm K
specific heat capacity of 0.208 J/gm K
The specific heat capacity is defined as the heat required to raise the temperature of a substance which is 1 gram. The temperature is in Kelvin and energy required is in joules.
Answer:
The answer to your question is 50 moles of O₂
Explanation:
Balanced Chemical reactions
1.- N₂(g) + 3H₂ (g) ⇒ 2NH₃ (g)
2.- 4NH₃ (g) + 5O₂(g) ⇒ 4NO (g) + 6H₂O (l)
moles of N₂(g) = 20 moles
moles of O₂(g) = ?
Process
1.- Calculate the moles of NH₃
1 mol of N₂ ------------- 2 moles of NH₃
20 moles of N₂ --------- x
x = (20 x 2) / 1
x = 40 moles of NH₃
2.- Calculate the moles of O₂
4 moles of NH₃ -------------- 5 O₂
40 moles of NH₃ ------------ x
x = (40 x 5) / 4
x = 200 / 4
x = 50 moles of O₂
Answer:
the correct answer is (sp3d2) (d)
Explanation:
The answer is 200 g.
If the molar mass of CaCl2 is 110.98 g/mol, this means there are 110.98 g in 1 L of 1 M solution.
Let's find how many g of CaCl2 are present in 0.720 M.
110.98 g : 1 M = x : 0.720 M
x = 110.98 g * 0.720 M : 1 M
x = 79.90 g
So there are 79.90 g in 0.720 M. In other words, in 1 L of 0.720 M solution there will be 79.90 g.
Now, we need to prepare ten beakers with 250 mL of solutions:
10 * 250 mL = 2500 mL = 2.5 L
79.90 g : 1 L = x : 2.5 L
x = 79.90 g * 2.5 L : 1 L
x = 199.75 g ≈ 200 g
