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3 years ago

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#1: Jim, Jane, Ann, and Bill measure an object’s length, density, mass, and volume, respectively. Which student’s measurement mi

ght be in centimeters?
A. Bill’s

B. Jane’s

C. Jim’s

D. Ann’s
**MY answer: C. Jim's

#2: How many centimeters are in 0.05 kilometers?

A. 50

B. 500

C. 5,000

D. 50,000
***My answer: C. 5,000

2 answers:

STatiana [176]3 years ago

7 0

1. The <span>student with a measurement that might be in centimeters is A. Bill.
2. C</span><span>entimeters in 0.05 kilometers is </span>C. 5,000
solution:
0.05 km x (1000 meters/ 1 km)
= 50 meters x (100 cm/ 1 meter)
=5000 cm

KiRa [710]3 years ago

4 0

Answer:

1. A. Bill.

2. C. 5,000

Explanation:

done the topic

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When 100.g Mg3N2 reacts with 75.0 g H2O, what is the maximum theoretical yield of NH3?

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Answer : The correct option is, 23.6 g

Explanation : Given,

Mass of $Mg_3N_2$ = 100.0 g

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Molar mass of $Mg_3N_2$ = 101 g/mol

Molar mass of $H_2O$ = 18 g/mol

First we have to calculate the moles of $Mg_3N_2$ and $H_2O$ .

$\text{Moles of }Mg_3N_2=\frac{\text{Given mass }Mg_3N_2}{\text{Molar mass }Mg_3N_2}$

$\text{Moles of }Mg_3N_2=\frac{100.0g}{101g/mol}=0.990mol$

and,

$\text{Moles of }H_2O=\frac{\text{Given mass }H_2O}{\text{Molar mass }H_2O}$

$\text{Moles of }H_2O=\frac{75.0g}{18g/mol}=4.17mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$Mg_3N_2(s)+6H_2O(l)\rightarrow 3Mg(OH)_2(aq)+2NH_3(g)$

From the balanced reaction we conclude that

As, 6 moles of $H_2O$ react with 1 mole of $Mg_3N_2$

So, 4.17 moles of $H_2O$ react with $\frac{4.17}{6}=0.695$ moles of $Mg_3N_2$

From this we conclude that, $Mg_3N_2$ is an excess reagent because the given moles are greater than the required moles and $H_2O$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NH_3$

From the reaction, we conclude that

As, 6 moles of $H_2O$ react to give 2 moles of $NH_3$

So, 4.17 moles of $H_2O$ react to give $\frac{2}{6}\times 4.17=1.39$ mole of $NH_3$

Now we have to calculate the mass of $NH_3$

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

Molar mass of $NH_3$ = 17 g/mole

$\text{ Mass of }NH_3=(1.39moles)\times (17g/mole)=23.6g$

Therefore, the maximum theoretical yield of $NH_3$ is, 23.6 grams.

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garri49 [273]

Answer:

Ka = 3.50x10⁻⁴

Explanation:

First, we need to convert the unit of 3.60 g/L to mol/L:

$C_{C_{9}H_{8}O_{4}} = 3.60 \frac{g}{L}*\frac{1 mol}{180.16 g} = 0.0200 mol/L$

The reaction dissociation of aspirin in water is:

C₉H₈O₄ + H₂O ⇄ C₉H₇O₄⁻ + H₃O⁺

0.02 - x x x

The constant of the above reaction is:

$Ka = \frac{[C_{9}H_{7}O_{4}^{-}][H_{3}O^{+}]}{[C_{9}H_{8}O_{4}]}$

$Ka = \frac{x^{2}}{0.02 - x}$

To find Ka we need to find the value of x. We know that pH = 2.6 so:

$pH = -log[H_{3}O^{+}]$

$2.6 = -log(x)$

$x = 2.51 \cdot 10^{-3} M = [H_{3}O^{+}] = [C_{9}H_{7}O_{4}^{-}]$

Now, the concentration of C₉H₈O₄ is:

$C_{C_{9}H_{8}O_{4}} = 0.02 - 2.51 \cdot 10^{-3} = 0.018 M$

Finally, Ka is:

$Ka = \frac{[C_{9}H_{7}O_{4}^{-}][H_{3}O^{+}]}{[C_{9}H_{8}O_{4}]} = \frac{(2.51 \cdot 10^{-3})^{2}}{0.018} = 3.50 \cdot 10^{-4}$

Therefore, the Ka of aspirin is 3.50x10⁻⁴.

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