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2 years ago

Select the atomic models that belong to the same element.

Chemistry

1 answer:

iogann1982 [59]2 years ago

8 0

Top left and bottom right are the atomic models of the same element

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A container has a volume of 2.79 L and a pressure of 5.97 atm. If the pressure changes to 1460 mm Hg, what is the container’s ne

maw [93]

Answer:

8.68 L is the new volume

Explanation:

You use Boyle's law for this.

$P_{1}V_{1}=P_{2}V_{2}$

$P_{1}$ = first pressure

$P_{2}$ = second pressure

$V_{1}$ = first volume

$V_{2}$ = second volume

Convert pressure from atm to mmHg (use same units):

5.97 x 760 = 4537.2 -> 4.54 x 10³

...maintain 3 significant figures in calculation, and round as needed...

(4.54 x 10³ mmHg)(2.79 L) = (1460 mmHg)( $V_{2}$ )

(4.54 x 10³ mmHg)(2.79 L) / (1460 mmHg) = $V_{2}$ = 8.68 L

Hope this helps :)

5 0

2 years ago

Give the nuclear symbol for the isotope of beryllium for which a=10? enter the nuclear symbol for the isotope (e.g., 42he).

galben [10]

An isotope is defined as an element that has the same number of protons as the common element but different number of neutrons. In this case, a beryllium atom has a molar weight of 10 amu. Thus, there are 4 protons and 6 neutrons. The nuclear symbol of Be-10 is 4 Be10

8 0

3 years ago

What is the relative humidity of the air when the dry-bulb temperature is 4°C and the dewpoint is -4°C?

boyakko [2]

Answer:56%

Explanation:

In the dewpoint chart when you line it up it ends up at 56%

3 0

3 years ago

What is the volume of 0.640 grams of Oz gas at Standard Temperature and Pressure (STP)?

Ilia_Sergeevich [38]

Answer: The volume of 0.640 grams of $O_{2}$ gas at Standard Temperature and Pressure (STP) is 0.449 L.

Explanation:

Given: Mass of $O_{2}$ gas = 0.640 g

Pressure = 1.0 atm

Temperature = 273 K

As number of moles is the mass of substance divided by its molar mass.

So, moles of $O_{2}$ (molar mass = 32.0 g/mol) is as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.640 g}{32.0 g/mol}\\= 0.02 mol$

Now, ideal gas equation is used to calculate the volume as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

$PV = nRT\\1.0 atm \times V = 0.02 mol \times 0.0821 L atm/mol K \times 273 K\\V = 0.449 L$

Thus, we can conclude that the volume of 0.640 grams of $O_{2}$ gas at Standard Temperature and Pressure (STP) is 0.449 L.

6 0

3 years ago

2 Points

djyliett [7]

A. Camouflage Fur is the correct answer.

7 0

2 years ago