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svetoff [14.1K]
3 years ago
5

What is the pressure of 20.5 mols of helium gas at 444 K and 999 L

Chemistry
1 answer:
Charra [1.4K]3 years ago
5 0

Answer:

IDK

Explanation:

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Which cellular process produces simple sugars?
Nataly_w [17]

Answer:

The answer is C. Hope this helps you out!

7 0
2 years ago
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Which is the best way to represent 0.0035 kg by using scientific notation?
Zolol [24]
The answer is 3.5 × 10^-3
7 0
3 years ago
Find the number of CoCl2 units present in a 0.78 mol sample.
stich3 [128]

\qquad\qquad\huge\underline{{\sf Answer}}

Here we go ~

1 mole of \sf{ COCl_2} has 6.022 × 10²³ molecules of the given compound.

So, 0.78 mole of \sf{COCl_2} will have ~

\qquad \sf  \dashrightarrow \: 0.78 \times 6.022 \times 10 {}^{23}

\qquad \sf  \dashrightarrow \:  \approx4.7 \times 10 {}^{23}  \:  \:  \: molecules

5 0
2 years ago
Determine the maximum number of electrons in the 2f designation
sergejj [24]
Google said

How many electrons fit in each shell around an atom?

The maximum number of electrons that can occupy a specific energy level can be found using the following formula:

Electron Capacity = 2n2

The variable n represents the Principal Quantum Number, the number of the energy level in question.

Energy Level
(Principal Quantum Number) Shell Letter Electron Capacity
1 K 2
2 L 8
3 M 18
4 N 32
5 O 50
6 P 72
Keep in mind that an energy level need not be completely filled before electrons begin to fill the next level. You should always use the Periodic Table of Elements to check an element's electron configuration table if you need to know exactly how many electrons are in each level.
4 0
3 years ago
Determine the concentration of a solution made by dissolving 44.0 g of calcium chloride (CaCl2) in 0.30 L of solution. SHOW YOUR
ycow [4]

Answer:

[CaCl₂] = 1.32 M

Explanation:

We know the volume of solution → 0.30 L

We know the mass of solute → 44 g of CaCl₂

Let's convert the mass of solute to moles.

44 g . 1 mol / 110.98 g = 0.396 moles

Molarity (mol/L) → 0.396 mol / 0.3 L  = 1.32 M

6 0
3 years ago
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