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polet [3.4K]
2 years ago
9

Label the following as a pure substance or a mixture.

Chemistry
2 answers:
Zigmanuir [339]2 years ago
6 0

Answer: a mixture

Explanation:

Snezhnost [94]2 years ago
5 0

Answer:

b

Explanation:

can you help ,meeeeee

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The diameter of a carbon atom is approximately 144 pm. Assuming a nanocontainer has a simple cubic structure and have an edge le
ivann1987 [24]
To answer this question, you need to understand the naming of meter unit system. The unit pm stands for pico meter which was 1/1000 of nanometer. Then, the calculation would be:

Number of atom= container length / atom diameter
Number of atom = 25 nm x (1000pm/nm) / 144pm= 173.6 atom
5 0
3 years ago
In the space, show a correct numerical setup for calculating the number of moles of CO2 present in 11 grams of CO2
makvit [3.9K]

Ans: Moles of CO2 = 0.25 moles

<u>Given:</u>

Mass of CO2 = 11 g

<u>To determine </u>

Moles of CO2

<u>Explanation:</u>

Moles = mass/molecular mass

Molecular mass of CO2 = 44 g/mol

Moles of CO2 = 11 g/44 g.mol-1 = 0.25 moles

4 0
3 years ago
Due to the fact that ionic compounds have strong intermolecular forces they are __________ at room temperature.
MrMuchimi
The correct answer should be solids
3 0
3 years ago
____ is involved in glucose metabolism.
krek1111 [17]

I took the test and the answer is chronium

7 0
3 years ago
For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction 2 NO ( g
enyata [817]

Answer:

ΔG = -61.5 kJ/mol (<u>Spontaneous process</u>)

Explanation:

2 NO (g)  +  O₂  (g)   ⇄  2NO₂ (g)

Let's apply the thermodynamic formula to calculate the ΔG

ΔG = ΔG° + R .T . lnQ

We don't know if the gases are at equilibrium, that's why we apply Q (reaction quotient)

ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln Q

How can we know Q? By the partial pressures (Qp)

P NO = 0.450atm

PO₂ = 0.1 atm

PNO₂ = 0.650 atm

Qp = [NO₂]² / [NO]² . [O₂]

Qp = 0.650² / 0.450² . 0.1 = 20.86

ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln 20.86

ΔG = -61.5 kJ/mol (<u>Spontaneous process</u>)

5 0
3 years ago
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