Explanation:
6 F2------->4 AlF3
F2-----------> 4/6 AlF3
8.25 F2 ---------> 4×8.25/6 AlF3
so 5.5 moles
Answer:
Q = 30284.88 j
Explanation:
Given data:
Mass of ethanol = 257 g
Cp = 2.4 j/g.°C
Chnage in temperature = ΔT = 49.1°C
Heat required = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Now we will put the values in formula.
Q = 257 g× 2.4 j/g.°C × 49.1 °C
Q = 30284.88 j
Answer:
Density = 11.4 g/cm³
Explanation:
Given data:
Density of lead = ?
Height of lead bar = 0.500 cm
Width of lead bar = 1.55 cm
Length of lead bar = 25.00 cm
Mass of lead bar = 220.9 g
Solution:
Density = mass/ volume
Volume of bar = length × width × height
Volume of bar = 25.00 cm × 1.55 cm × 0.500 cm
Volume of bar = 19.4 cm³
Density of bar:
Density = 220.9 g/ 19.4 cm³
Density = 11.4 g/cm³
The HCl added = 1.25 moles
and the moles of Na2HPO4 = 1 mole
Now when acid is added in the given solution of Na2HPO4
One mole of H+ will react with one mole of Na2HPO4 to given one mole of NaH2PO4
Na2HPO4 + H+ ---> NaH2PO4
Now this one mole formed NaH2PO4 will further react with 0.25 moles of H+ left to form 0.25 moles of H3PO4 and 0.75 moles of NaH2PO4 will remain in the solution
So this will result into formation of a buffer of phosphoric acid and NaH2PO4
NaH2PO4 + H+ ---> H3PO4
pKa of H3PO4 = 2.1
so pH = pKa + log [salt] / [acid] = 2.1 + log [0.75 / 0.25] = 2.58
so the pH will be in between 2.1 to 7.2
Answer:
moenkopi formation because layers further down are always older. think about it as a pile of laundry the clothes at the bottom of the pile were worn earlier in the week and are older and dirtier.
Explanation:
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