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3 years ago

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The dimensions of a box are measured to be 18.4 inches by 17.92 inches by 26 inches. The volume of the box can be found by multi

plying these three dimensions. What is the volume of the box expressed to the correct number of significant figures?

2 answers:

fiasKO [112]3 years ago

5 0

The answer is 8572.93 in.

madam [21]3 years ago

4 0

The volume of the box is 8.6 × 10³ in².

<em>V = lwh</em> = 26 in × 18.4 in × 17.92 in = 8.6 × 10³ in²

<em>Note</em>: The answer can have only <em>two significant figures</em> because that is all you gave for the length of the box.

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A gray element that borders on the zigzag line of the periodic table, is ductile and malleable, but is not a very good conductor

frez [133]

The answer is non metal D

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2 years ago

What causes valley air to move up surrounding mountains during the day?

AleksandrR [38]

Answer:

so the third one

Explanation:

During the day, the sun heats up mountain air rapidly while the valley remains relatively cooler. Convection causes it to rise, causing a valley breeze. At night, the process is reversed. During the night the slopes get cooled and the dense air descends into the valley as the mountain wind.

7 0

2 years ago

Read 2 more answers

If a red blood cell is 100% saturated, how many molecules of O2 are bound to it?1 billion molecules of O24 molecules of O2250 mi

kondor19780726 [428]

Answer:

1 billion molecules O₂

Explanation:

From my research, a human red blood cell contains approximately 270 million hemoglobin molecules.

A hemoglobin molecule contains four heme groups, <em>each of which has an iron ion forming a coordination complex that carries every dioxygen molecule. </em>Therefore for each hemoglobin molecule, we will have 4 dioxygen molecules. The heme groups are responsible for the transport of every dioxygen and other diatomic gases.

Hence, the number of O₂ molecules in a red blood cell saturated with 100% will be:

$\frac{270 \cdot10^{6} hemoglobine molecules}{1 red blood cell} \cdot \frac{4 heme group}{1 hemoglobine molecule} \cdot \frac{1 O_{2} molecules}{1 heme group} = 1 \cdot 10^{9} O_{2} \frac{molecules}{red blood cell}$

So, the correct answer is 1 billion of O₂ molecules.

Have a nice day!

8 0

3 years ago

The pressure in a car tire is 198 kPa at 27°C. After a long drive, the pressure

AlekseyPX

Answer : The temperature of the air in the tire is, 341 K

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

$P\propto T$

or,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1$ = initial pressure = 198 kPa

$P_2$ = final pressure = 225 kPa

$T_1$ = initial temperature = $27^oC=273+27=300K$

$T_2$ = final temperature = ?

Now put all the given values in the above equation, we get:

$\frac{198kPa}{300K}=\frac{225kPa}{T_2}$

$T_2=340.9K\approx 341K$

Therefore, the temperature of the air in the tire is, 341 K

5 0

3 years ago

Compute the percent ionic character of the interatomic bonds for the following compounds : a. TiO2 b. ZnTe c. CsCld. InSb e. MgC

sesenic [268]

Answer:

a. 63.2%

b. 11.7%

c. 73.3%

d. 0.995%

e. 55.5%

Explanation:

An ionic compound is a compound that is formed by ions, so one of the elements must donate electrons (which is the cation, the positive ion), and the other will receive these electrons (which is the anion, the negative ion).

The power of an element has to attract the electrons is called electronegativity, and so, as higher is the difference of electronegative of the elements, it is more probable that one of them will "still" the electrons and will form an ionic compound. The percent of this ionic character can be found by the Pauling's equation:

$%IC = (1 - e^{-0.25*(x_A - x_B)^2})$ *100%

Where $x_A - x_B$ is the electronegativity difference of the elements. Thus, consulting an electronegativity table:

a. $x_{Ti}$ = 1.5

$x_{O}$ = 3.5

$%IC = (1 - e^{-0.25*(3.5 - 1.5)^2})$ *100%

%IC = 63.2%

b. $x_{Zn}$ = 1.6

$x_{Te}$ = 2.1

$%IC = (1 - e^{-0.25*(2.1 - 1.6)^2})$ *100%

%IC = 11.7%

c. $x_{Cs}$ = 0.7

$x_{Cl}$ = 3.0

$%IC = (1 - e^{-0.25*(3.0 - 0.7)^2})$ *100%

%IC = 73.3%

d. $x_{In}$ = 1.7

$x_{Sb}$ = 1.9

$%IC = (1 - e^{-0.25*(1.9 - 1.7)^2})$ *100%

%IC = 0.995 %

e. $x_{Mg}$ = 1.2

$x_{Cl}$ = 3.0

$%IC = (1 - e^{-0.25*(3.0 - 1.2)^2})$ *100%

%IC = 55.5%

4 0

3 years ago