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Troyanec [42]
2 years ago
12

HELP ME! If I get an F on my test I’m getting kicked out:(

Chemistry
1 answer:
satela [25.4K]2 years ago
3 0

Answer:

It is just slightly less abundant than its alkali cousin, sodium. Potassium is less dense than water, so it can float on water. However, chemically, potassium reacts with water violently. It will give off hydrogen and eventually catch fire.

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Helga [31]
Metallic I’m pretty sure. :)
4 0
3 years ago
Which pair of elements can be used to determine the age of a fossil that is over one billion years old?
lakkis [162]

Answer:

Explanation:

There are some radioactive nuclides can be used to measure time on an archeological scale. One is the best example of this is radiocarbon dating. This process is based on the ratio of caebon-14 to carbon-12 in the atmosphere which is relatively constant.

The half time of C-14 5730 years  

Carbon-14 is a radioactive nucleus. It has a half-life of 5730 years.

All living tissues like plants and animal absorbed carbon-12 along with carbon-14 with same ratio of caebon-14 to carbon-12 in the atmosphere.

Carbon-14 dating is based on the ratio of carbon-14 to carbon-12 in the atmosphere which is relatively constant

3 0
3 years ago
Read 3 more answers
Standard reduction half-cell potentials at 25∘c half-reaction e∘ (v) half-reaction e∘ (v) au3 (aq) 3e−→au(s) 1.50 fe2 (aq) 2e−→f
Zanzabum

<em>K</em> = 5.0 × 10^25

<h2>Part (a). Calculate <em>E</em>° for the reaction </h2>

<em>Step 1.</em> Write the equations for the two half-reactions

2H^(+)(aq) + 2e^(-) → H2(g); _0.00 V

Zn^(2+)(aq) + 2e^(-) → Zn(s); -0.76 V

<em>Step 2.</em> Identify the cathode and the anode

The half-cell with the more negative <em>E</em>° (Zn) is the anode.

<em>Step 3.</em> Calculate <em>E</em>°

Zn(s) → Zn^(2+)(aq) + 2e^(-); _________+0.76 V

2H^(+)(aq) + 2e^(-) → H2(g); __________0.00 V

Zn(s) + 2H^(+)(aq) → Zn^(2+)(aq) + H2(g); +0.76 V

<em>E</em>° = +0.76 V

<h2>Part (b). Calculate <em>K</em> for the reaction </h2>

The relation between <em>E</em>° and <em>K</em> is

<em>E</em>° = (<em>RT</em>)/(<em>nF</em>)ln<em>K </em>

where

<em>R</em> = the universal gas constant: 8.314 J·K^(-1)mol^(-1)

<em>T</em> = the Kelvin temperature

<em>n</em> = the moles of electrons transferred

<em>F</em> = the Faraday constant: 96 485 J·V^(-1)mol^(-1)

Then

0.76 V = [8.314 J·K^(-1)mol^(-1) × 298.15 K]/[2 × 96 485 J·V^(-1)mol^(-1)]ln<em>K</em>

0.76 = 0.012 85 ln<em>K</em>

ln<em>K</em> = 0.76/0.012 85 = 59.16

<em>K</em> =e^59.16 = 5.0 × 10^25

4 0
3 years ago
To what temperature must a balloon, initially at 25°c and 2.00 l, be heated in order to have a volume of 6.00 l?
Kaylis [27]
V1 = 2.00 L 
<span>T1 = 25 + 273 = 298 K </span>
<span>V2 = 6.00 L </span>
<span>T2 = ? </span>
<span>Assuming the pressure is to remain constant, then </span>
<span>V1/T1 = V2/T2 </span>
<span>T2 = T1V2/V1 = (298)(6)/(2) = 894 deg K</span>
6 0
3 years ago
Read 2 more answers
What is the volume of these gases atSTP A-3.20x10^-3mol CO2
pickupchik [31]
I dont know the answer sorry follow me for a follow back
4 0
3 years ago
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