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Eduardwww [97]
4 years ago
8

I have very good ability to conduct electricity. I am never found alorie in nature. When I combine with other elements, I usuall

y give up my one valence electron. I am the only element in my group with a one-letter symbol.
Chemistry
1 answer:
goldenfox [79]4 years ago
3 0

Answer:

Potassium

Explanation:

The description here fits potassium. Potassium is not found alone in nature as it is very reactive. Hence, it tends to occur in relation with other elements because of its high reactivity.

Because it is stated that it gives up its single valence electron, this shows that it belongs to group one of the periodic table. We might be tempted to think hydrogen is the correct answer. But it is interesting to note that hydrogen is not even supposed to be placed in group one at all. It is only placed there for convenience

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What is the volume at STP of 3.44 x 1023 molecules of CO2
almond37 [142]

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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
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