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AfilCa [17]
2 years ago
7

Fredrick is doing a lab on pH. He uses many household items to determine the pH of his items. Which of his actions best describe

s inappropriate lab safety?
He changes into closed-toed shoes before lab starts.
He doesn’t wear goggles because he is using household chemicals.
He asks his teacher if he can mix high- and low-pH solutions before discarding them.
He leaves his workstation to identify all of the safety equipment in the lab.
Chemistry
2 answers:
Oksana_A [137]2 years ago
4 0

The correct answer is B. He doesn’t wear goggles because he is using household chemicals.

Explanation:

Carrying out experiments especially those that imply substances such as chemicals implies following practices to guarantee safety. Because of this, in the lab you should use special clothes this includes goggles and gloves; you should clean and decontaminate your station; avoid eating, drinking or smoking and always ask your teacher for any step or procedure. This implies, the inappropriate action is "He doesn’t wear goggles because he is using household chemicals", because protection should be used all the time as even substance we use every day can be dangerous if you are closely exposed to them or if they are mixed with others.

NNADVOKAT [17]2 years ago
3 0

The correct answer is A.

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Answer:

\large \boxed{\text{1.22 mg/s}}

Explanation:

We can use the Noyes-Whitney equation to calculate the rate of dissolution.

\dfrac{\text{d}M}{\text{d}t} = \dfrac{DA(C_{s} - C)}{d}

Data:

D = 1.75 × 10⁻⁷ cm²s⁻¹

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Cₛ - C = (0.35 - 2.1 × 10⁻⁴) mg·cm⁻³ = 0.350 mg·cm⁻³

d = 1.25 µm = 1.25 × 10⁻⁶ m = 1.25 × 10⁻⁴ cm

\dfrac{\text{d}M}{\text{d}t} = \dfrac{(1.75 \times 10^{-7} \text{cm}^{2}\text{s}^{-1})(2.5 \times 10^{3} \text{ cm}^{2})(0.350\text{ mg$\cdot$cm$^{-3}$})}{1.25 \times 10^{-4} \text{ cm}} = \textbf{1.22 mg/s}\\\\\text{The rate of dissolution is $\large \boxed{\textbf{1.22 mg/s}}$}

8 0
2 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

Best regards!

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