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The total of Konrad's purchases ledger balances was $57,400. Later, these errors were found:
| Discount allowed overcast in cash book | $2,000 |
| Returns outwards omitted in a supplier's account | $350 |
| Payments to trade payables undercast in cash book | $137 |
| Purchases journal overcast | $500 |
What is the corrected total of the trade payables balances?
Only errors affecting individual supplier accounts change the purchases ledger total.
1. Returns outwards omitted: $57,400 - $350 = $57,050.
The others (Discount allowed, payments undercast in cash book, journal overcast) affect the Control Account or other ledgers, but not the individual ledger balances until corrected there.
Calculation of oxidation states:
- $[Cr(C_{6}H_{6})_{2}]$: Benzene is a neutral ligand, so $Cr$ is in $0$ oxidation state.
- $[Cr(H_{2}O)_{6}]Cl_{3}$: $H_{2}O$ is neutral and there are 3 $Cl^{-}$ ions, so $x + 0 = +3 \Rightarrow x = +3$.
- $K_{2}[Cr(CN)_{2}(O)_{2}(O_{2})(NH_{3})]$: $K$ is $+1$, $CN$ is $-1$, $O$ (oxo) is $-2$, $O_{2}$ (peroxo) is $-2$, and $NH_{3}$ is $0$. $2(+1) + x + 2(-1) + 2(-2) + (-2) + 0 = 0 \Rightarrow x = +6$.
If $n = 2^3$, what is the value of $n^n$?
$n = 2^3 = 8$.
$n^n = 8^8 = (2^3)^8 = 2^{3 \times 8} = 2^{24}$.
Answer: $\mathbf{2^{24}}$.
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