Does anyone have the answers to the end of semester test B for chemistry on Edmentum?

Chemistry

1 answer:

Amanda [17]3 years ago

6 0

No I do not have the answers

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Aluminum has a density of 2.7 g/cm3, how much space in cm3 would 81 grams of aluminum occupy? Show steps to answering this equat

Evgesh-ka [11]

Answer:

30 cm³

Explanation:

Step 1: Given data

Density of aluminum (ρ): 2.7 g/cm³

Mass of aluminum (m): 81 g

Volume occupied by aluminum (V): ?

Step 2: Calculate the volume occupied by aluminum

The density of aluminum is equal to its mass divided by its volume.

ρ = m/V

V = m/ρ

V = 81 g / 2.7 g/cm³

V = 30 cm³

4 0

3 years ago

A student heats 10.52 g of sodium hydrogen carbonate in a crucible until the compound completely decomposes to sodium carbonate

saveliy_v [14]

Answer:

$m_{Na_2CO_3}^{theoretical}=6.636gNa_2CO_3$

Explanation:

Hello!

In this case, since the decomposition of sodium hydrogen carbonate is:

$2NaHCO_3(s)\rightarrow Na_2CO_3(s)+H_2O(g)+CO_2(g)$

Thus, since there is a 2:1 mole ratio between the sodium hydrogen carbonate and sodium carbonate, and the molar masses are 84.01 and 105.99 g/mol respectively, we obtain the following theoretical yield:

$m_{Na_2CO_3}^{theoretical}=10.52gNaHCO_3*\frac{1molNaHCO_3}{84.01gNaHCO_3}*\frac{1molNa_2CO_3}{2molNaHCO_3} *\frac{105.99gNa_2CO_3}{1molNa_2CO_3}\\\\ m_{Na_2CO_3}^{theoretical}=6.636gNa_2CO_3$