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3 years ago

Look at the picture below. Helppp

Chemistry

1 answer:

kykrilka [37]3 years ago

7 0

Group 2 is alkali earth metals
group 17 is halogens
group 18 is noble gasses
group 1 is alkali metals

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What is the volume at STP of 3.44 x 1023 molecules of CO2

almond37 [142]

Answer:

C. 12.8 liters.

Explanation:

The Standard Temperature and Pressure (STP) of a gas are 273.15 K and 100 kilopascals. From Avogadro's Law, a mole of carbon dioxide contains $6.022 \times 10^{23}$ molecules. If we suppose that carbon dioxide behaves ideally, then the equation of state for ideal gas is:

$P\cdot V = n\cdot R_{u}\cdot T$ (1)

$P\cdot V = \frac{r\cdot R_{u}\cdot T}{N_{A}}$ (1b)

Where:

$P$ - Pressure, measured in pascals.

$V$ - Volume, measured in liters.

$r$ - Amount of molecules, no unit.

$N_{A}$ - Avogadro's number, no unit.

$R_{u}$ - Ideal gas constant, measured in pascal-liters per mole-Kelvin.

$T$ - Temperature, measured in Kelvin.

If we know that $P = 100000\,Pa$ , $r = 3.44\times 10^{23}$ , $N_{A} = 6.022\times 10^{23}$ , $T = 273.15\,K$ and $R_{u} = 8.314\times 10^{3}\,\frac{L\cdot Pa}{mol\cdot K}$ , then the volume of carbon dioxide at STP is:

$V = \frac{r\cdot R_{u}\cdot T}{N_{A}\cdot P}$

$V = \frac{(3.44\times 10^{23})\cdot \left(8.314\times 10^{3}\,\frac{L\cdot Pa}{mol\cdot K} \right)\cdot (273.15\,K)}{(6.022\times 10^{23})\cdot (100000\,Pa)}$

$V = 12.972\,L$

Therefore, the correct answer is C.

8 0

3 years ago

Where else have you heard the term conversation? How does that term’s meaning in that other context relate to its meaning in che

Mrac [35]

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4 years ago

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Maurinko [17]

Answer: what are the answers to the questions

Explanation:

5 0

3 years ago

An element has five isotopes. Calculate the atomic mass of this element using the information below. Show all your work. Using t

Katena32 [7]

Answer: Sol:-

Data provided in the question is :-

Atomic mass of isotope -1 = 64 amu

Atomic mass of isotope -2 = 66 amu

Atomic mass of isotope -3 = 67 amu

Atomic mass of isotope -4 = 68 amu

Atomic mass of isotope - 5 = 70 amu

Percentage abundace of isotope - 1 = 48.89 %

Percentage abundance of isotope -2 = 27.81 %

Percentage abundance of isotope - 3 = 4.11%

Percentage abundance of isotope-4 = 18.57%

Percentage abundance of isotope - 5 = 0.62 %

Formula used :-

Average atomic mass of an element =[ {(atomic mass of isotope-1 * percentage abundance of isotope-1) + ( atomic mass of isotope-2 * percentage abundance of isotope -2) + ( atomic mass of isotope -3 * percantege abundance of isotope-3 ) + ( atomic mass of isotope-4 * percentage abundance of isotope-4) + (atomic mass of isotope-5 * percentage abundance of isotope-5)} / 100]

Calculation :-

Put all the value in the formula :-

Average atomic mass of an element = [{(64 * 48.89) + (66 * 27.81) + (67 * 4.11) + (68 * 18.57) + (70 * 0.62)} / 100] amu

= [{(3128.96) + (1835.46) +(257.37) + (1262.76) + (43.4)} / 100] amu

= {(6528.04) / 100} amu

= 65.2804 amu

Average atomic mass of an element is = 65.2804 amu

Then this mass is approximatly equal to atomic mass of zinc so this element would be zinc

atomic mass of zinc = 65.38 \approx 65.2804 amu

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3 years ago

The Law of Conservation of Mass states that ____

Vaselesa [24]

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