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3 years ago

12

Three students use a meterstick with millimeter markings to measure a length of wire. Their measurements are 3 cm, 3.3 cm, and 2

.87 cm,
respectively. Explain which answer was recorded correctly.

2 answers:

Volgvan3 years ago

7 0

Answer: 3 cm

Explanation:

andreev551 [17]3 years ago

3 0

The answer is 3 I had this two

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What happens when

jolli1 [7]

Answer:

a i think but idk

Explanation:

3 0

3 years ago

A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor

AlexFokin [52]

Answer : The partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of $C_2HBrClF_3$ = 15.0 g

Mass of $O_2$ = 22.6 g

Molar mass of $C_2HBrClF_3$ = 197.4 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_2HBrClF_3$ and $O_2$ .

$\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole$

and,

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole$

Now we have to calculate the mole fraction of $C_2HBrClF_3$ and $O_2$ .

$\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971$

and,

$\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903$

Now we have to partial pressure of $C_2HBrClF_3$ and $O_2$ .

According to the Raoult's law,

$p^o=X\times p_T$

where,

$p^o$ = partial pressure of gas

$p_T$ = total pressure of gas

$X$ = mole fraction of gas

$p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T$

$p_{C_2HBrClF_3}=0.0971\times 862torr=84torr$

and,

$p_{O_2}=X_{O_2}\times p_T$

$p_{O_2}=0.903\times 862torr=778torr$

Therefore, the partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

6 0

3 years ago

According to the aufbau principle,_____.

Softa [21]

Answer:

Hi there!

The correct answer to this question is: electrons enter orbitals of lowest energy first.

5 0

3 years ago

Read 2 more answers

To convert from °F to °C: T(°C) = T(°F - 32) × 5/9

DaniilM [7]

Answer:

Your notation is a bit confusing, let me write it more clearly.

Explanation:

( Temperature in °F − 32) × 5/9 = Temperature in °C

4 0

3 years ago

Calculate the amount of HCN that gives the lethal dose in a small laboratory room measuring 12.0 ft×15.0 ft×8.60ft.

garik1379 [7]

<u>Given:</u>

Dimensions of the room= 12 ft * 15 ft * 8.60 ft

<u>To determine:</u>

The amount of HCN that gives the lethal dose in the room with the given dimensions

<u>Explanation:</u>

As per the World Health Organization, the lethal dose of HCN is around 300 ppm

300 ppm = 300 mg of HCN/ kg of inhaled air

Volume of air = volume of room = 12 * 15 *8.6 = 1548 ft³

Now, 1 ft³ = 28316.8 cm³

Therefore, the calculated volume of air corresponds to:

1548 * 28316.8 = 4.383 * 10⁷ cm3

Density of air (at room temperature 25 C) = 0.00118 g/cm3

Thus mass of air corresponding to the calculated volume is

Mass = Density * volume = 0.00118 g/cm3 * 4.383 * 10⁷ cm3

= 5.172*10⁴ g = 51.72 kg

Lethal amount of HCN corresponding to 51.72 kg of air would be.

= 51.72 kg air* 300 mg of HCN/1 kg air = 15516 mg

Ans: Lethal dose of HCN = 15.5 g

7 0

3 years ago