False: to protect iron from corrosion, keep it near moistureband oxygen.
Answer:
99.24%.
Explanation:
- NaCl reacted with AgNO₃ as in the balanced equation:
<em>NaCl + AgNO₃ → AgCl(↓) + NaNO₃,</em>
1.0 mol of NaCl reacts with 1.0 mol of AgNO₃ to produce 1.0 mol of AgCl and 1.0 mol of NaNO₃.
- We need to calculate the no. of moles of AgCl produced:
no. of moles of AgCl = mass/molar mass = (2.044 g)/(143.32 g/mol) = 0.0143 mol.
- Now, we can calculate the no. of moles of NaCl that can precipitated as AgCl (0.0143 mol), these moles represents the no. of moles of pure NaCl in the sample:
<em>using cross multiplication:</em>
1.0 mol of NaCl produce → 1.0 mol of AgCl, from the stichiometry.
∴ 0.0143 mol of NaCl produce → 0.0143 mol of AgCl.
- Now, we can get the mass of puree NaCl in the sample:
mass of pure NaCl = (no. of moles of pure NaCl)(molar mass of NaCl) = (0.0143 mol)(58.44 g/mol) = 0.8357 g.
∴ The percentage of NaCl in the impure sample = [(mass of pure NaCl)/(mass of the impure sample)] x 100 = [(0.8357 g)/(0.8421 g)] x 100 = 99.24%.
Answer:
16 °C
Explanation:
Step 1: Given data
- Provided heat (Q): 811.68 J
- Mass of the metal (m): 95 g
- Specific heat capacity of the metal (c): 0.534 J/g.°C
Step 2: Calculate the temperature change (ΔT) experienced by the metal
We will use the following expression.
Q = c × m × ΔT
ΔT = Q/c × m
ΔT = 811.68 J/(0.534 J/g.°C) × 95 g = 16 °C
Every atom tends to form configuration of noble gas , with the 8 electrons in valence shell.
Answer:
1.428 moles
Explanation:
If 0.0714 moles of N2 gas occupies 1.25 L space,
how many moles of N2 have a volume of 25.0 L?
Assume temperature and pressure stayed constant.
we experience it 0.0714 moles: 1.25L space
x moles : 25L of space
to get the x moles, cross multiply
(0.0714 x 25)/1.25
1.785/1.25 = 1.428 moles
