Answer:
0.57 moles (NH4)3PO4 (2 sig. figs.)
Explanation:
To quote, J.R.
"Note: liquid ammonia (NH3) is actually aqueous ammonium hydroxide (NH4OH) because NH3 + H2O -> NH4OH.
H3PO4(aq) + 3NH4OH(aq) ==> (NH4)3PO4 + 3H2O
Assuming that H3PO4 is not limiting, i.e. it is present in excess
1.7 mol NH4OH x 1 mole (NH4)3PO4/3 moles NH4OH = 0.567 moles = 0.57 moles (NH4)3PO4 (2 sig. figs.)"
the answer is <span>Astatine is one period further than tellurium, meaning it has an extra shell. Therefore, the At atom will be bigger than the Te atom. </span>
Answer:
2370.0 contains 4 significant digits and Option (c) is correct .
1.20\times 10^{-3}\ contains\ three\ significant\ digit.
Option (b) is correct .
Step-by-step explanation:
Rules for finding significant digit .
1 : Non-zero digits are always significant.
2: Any zeros between two significant digits are significant .
3: Trailing zeros in the decimal number is also significant.
As the number given be 2,370.0.
= \frac{23700}{10}
Simplify the above
= 2370
Thus by using the rule given above.
2370.0 contains 4 significant digits.
Option (c) is correct .
As the number given be 0.00120 .
= \frac{120}{100000}
Simplify the above
= \frac{1.20}{1000}
= 1.20\times 10^{-3}
Thus by using the rule given above.
1.20\times 10^{-3}\ contains\ three\ significant\ digit.
Option (b) is correct .
Answer:
2.6×10^-19 J/photon
Explanation:
E of photon = h × ν
where h= 6.63 × 10^-34 j.s
v= C ÷ λ
E = ( h × c) ÷ λ
E = (6.63 × 10^-34 × 3.00
×10^8 ) ÷ ( 765 × 10^-9)
E = 2.6×10^-19 J/photon
Answer:
option c the relative number of atoms of each element using the lowest ratio because empirical formula gives the simplest-whole number ration of each element
e.g. the empirical formula of benzeneC6H6 in CH
Explanation:
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