Lithium fluoride is a solid at room temperature because it is a salt that is held together by ionic bonds. Lithium. fluoride has a giant ionic structure.
C.) It was Einstein who created quantum of light
Answer
Na OH reacts with H Cl and forms Na Cl and H₂O
NaOH + HCl → NaCl + H₂O
Here we can see that 1 mole of NaOH reacting with 1 mole of HCl and forming 1 mole of NaCl and 1 mole of H₂O
when NaOH and HCl are added together in equal amount then they will completely neutralize each other but NaOH is hygroscopic in nature which means it can absorb water from air so it will not be weighted accurately.
hence, for neutralization we will take extra NaOH.
The empirical formula for pyrite is FeS2.
HOW TO CALCULATE EMPIRICAL FORMULA:
- The empirical formula represents the simplest whole number ratio of constituents element of a compound. The empirical formula of pyrite can be calculated as follows:
46.5 mass % Fe = 46.5g of Fe
53.5 mass % S = 53.5g of S
- Next, we divide each element's mass value by its molar mass
Fe = 46.5g ÷ 56g/mol = 0.83mol
S = 53.5g ÷ 32g/mol = 1.67mol
- Next, we divide each mole value by the smallest (0.83mol)
Fe = 0.83mol ÷ 0.83 = 1
S = 1.67mol ÷ 0.83 = 2.014
Approximately, the ratio of Fe to S is 1:2. Therefore, the empirical formula of pyrite is FeS2.
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Answer:
6626 g
Explanation:
Given that:
Density of water = 1.00 g/ml, volume of water = 42800 ml.
Since density = mass/ volume
mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g
Initial temperature of water = 22°C and final temperature of water = 45°C.
specific heat capacity for water = 4.184 J/g°C
ΔT water = 45 - 22 = 23°C
For iron:
mass = m,
specific heat capacity for iron = 0.444 J/g°C
Initial temperature of iron = 1445°C and final temperature of water = 45°C.
ΔT iron = 45 - 1445 = -1400°C
Quantity of heat (Q) to raised the temperature of a body is given as:
Q = mCΔT
The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.
Q water (gain) + Q iron (loss) = 0
Q water = - Q iron
42800 g × 4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C
m = 4118729.6/621.6
m = 6626 g