Answer: 40 grams
Explanation:
The quantity of Heat Energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since Q = 93.4J
M = ?
C = 0.129 J/g.C
Φ = 40.4°C - 22.3°C = 18.1°C
Then, Q = MCΦ
Make Mass, M the subject formula
M = Q/CΦ
M = (93.4J) / (0.129 J/g.C x 18.1°C)
M = 93.4J / 2.33J/g
M = 40 g
Thus, the mass of the lead is 40 grams
Answer:
Hi there!
I believe you are missing an attachment to this question however
I strongly believe that the answer you are looking for is 85.
Explanation:
If you provide the graph, the rate of strokes in Asian women will be 17 per 1,000 women so all you have to do is multiply 17 by 5 and you get 85
Answer: 17.34 grams of alum will be produced if 0.9875 g of Aluminium foil was used.
Explanation: Reaction to form alum from Aluminium is given as:
We are given Aluminium to be the limiting reactant, so the formation of alum will be dependent on Aluminium because it limits the formation of product.
By stoichiometry,
2 moles of Al is producing 2 moles of Alum
Mass of 2 moles of Aluminium = (2 × 27)g/mol = 54 g/mol
Mass of 2 moles of alum = (2 × 474)g/mol = 948 g/mol
54 g/mol of aluminium will produce 948 g/mol of alum, so
Amount of Alum produced = 17.34 grams
Theoretical yield of alum = 17.34 grams.
Answer:
it form a negative ion because the no. of negative charges exceeds that of the positive charges
1. For the first question, we must find the mass of the anhydrous salt, MgSO₄. The molar mass for MgSO₄·7H₂O is 246.47 g/mol, while that of MgSO₄ is 120.37 g/mol.
7.834 g MgSO₄·7H₂O * 1 mol MgSO₄·7H₂O/246.47 g * 1 mol MgSO₄/1 mol MgSO₄·7H₂O * 20.37 g MgSO₄/mol = <em>0.647 g MgSO₄</em>
2. Mass of Water = Mass of sample - Mass of anhydrous sale
Mass of water = 7.834 - 0.647 = 7.19 g
Percent Water in Hydrate = 7.19/7.834 * 100 = <em>91.74%</em>
