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aniked [119]
3 years ago
8

Please help me

Chemistry
1 answer:
natta225 [31]3 years ago
4 0
Indefinite shape and definite volume I’m pretty sure I could be wrong tho
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If a 22.4 L volume of a sample of gas has a density of 0.900 grams/L at 1.00 atm and 0.00°C. Given
disa [49]

Answer:

Formula Weight of gas sample = 20.1 g/mole => Neon (Ne)

Explanation:

Use Ideal Gas Law formula to determine formula weight and compare to formula weights of answer choices.

PV = nRT = (mass/fwt)RT => fwt = (mass/Volume)RT = Density x R x T

Density = 0.900 grams/L

R = 0.08206 L·atm/mole·K

T = 0.00°C = 273Kfwt = (0.900g/L)(0.08206L·atm/mole·K )(273K)

= 20.1 g/mol => Neon (Ne)

4 0
3 years ago
A solution that contains the maximum amount of solute that is soluble at a given temperature is said to be —
solniwko [45]

Answer: Saturated

A solution that contains the maximum amount of solute that is soluble at a given temperature is said to be saturated.

Explanation:

A Saturated solution is one that contains as much (i.e maximum) solute as it can dissolve at that temperature in the presence of undissolved solute particles.

For instance: if a given volume of water can only dissolve a certain amount of salt in it at room temperature, then, more salt added will not dissolve.

Thus, making the solution saturated.

5 0
3 years ago
How can acid rain affect decomposers in an ecosystem?
adell [148]

Answer:

I think it would be C or D

Explanation:

I'm sorry I couldn't give you the correct answer, but glad I was able to try and help you. Have a blessed day!

4 0
3 years ago
Which statement provides background knowledge that helps a reader understand the theme of valuing all life found in"Birdfoot's G
alekssr [168]

Answer:

i think its A

Explanation:

7 0
4 years ago
Read 2 more answers
THIS IS DUE TODAY!!!!!!!!!!!!!!!!!!!!!!!!!! PLEASE HELP ME ASAPPPPPPPP CHEMISTRY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
allochka39001 [22]

Answer:

\large \boxed{\text{0.1767 mol/L}}$

Explanation:

(a) Balanced equation

3NaOH + H₃PO₄ ⟶ Na₃PO₄ + 3H₂O  

(b) Moles of NaOH

\text{Moles of  NaOH} = \text{34.52 mL NaOH} \times \dfrac{\text{0.3840 mmol NaOH}}{\text{1 mL NaOH}} = \text{13.26 mmol NaOH}

(c) Moles of H₃PO₄

The molar ratio is 1 mol H₃PO₄:3 mol NaOH.

\text{Moles of H$_{3}$PO}_{4} = \text{13.26 mmol NaOH} \times\dfrac{\text{ 1 mmol H$_{3}$PO}_{4}}{\text{3 mmol NaOH}}\\\\= \text{4.419 mmol H$_{3}$PO}_{4}

(d) Molar concentration of H₃PO₄

c = \dfrac{\text{moles of solute}}{\text{litres of solution}}\\\\c = \dfrac{n}{V} = \dfrac{\text{4.419 mmol}}{\text{25.00 mL}} = \text{0.1767 mol $\cdot$ L$^{-1}$}\\\\\text{The molar concentration of the NaOH is $\large \boxed{\textbf{0.1767 mol/L}}$}

7 0
3 years ago
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