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zheka24 [161]
3 years ago
5

Why potasium more reactive than lithium and sodium

Chemistry
1 answer:
frez [133]3 years ago
5 0
So potassium is more reactive than lithium because the outer electron of a potassium atom is further from its nucleus than the outer electron of a lithium atom. Hope this answers the question. Have a nice day. Feel free to ask more questions.
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What kinds of places may hold life ?
emmasim [6.3K]

The planets that could host liquid water and might have rocky ... Liquid water is a necessity for the kind of life we know

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3 years ago
The formation of tert-butanol is described by the following chemical equation:
fomenos

Answer:

Second step

(CH3)3C+ (aq) + OH^-(aq) ------->(CH3)3COH(aq)

Explanation:

This reaction involves;

First the ionization of the tertiary halide to firm a carbocation

Secondly the attack of the hydroxide ion on the carbocation to form tert-butanol

First step;

(CH3)3CBr (aq) → (CH3)3C+ (aq) + Br- (aq)

Second step

(CH3)3C+ (aq) + OH^-(aq) ------->(CH3)3COH(aq)

This second step completes the reaction mechanism.

3 0
3 years ago
What is the Newton's fourth law​
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Answer:

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6 0
3 years ago
Read 2 more answers
Consider the following reaction where Kc = 1.29×10-2 at 600 K: COCl2 (g) CO (g) + Cl2 (g) A reaction mixture was found to contai
vaieri [72.5K]

Answer:

1. In order to reach equilibrium COCl₂(g) must be consumed.

B. False

2. In order to reach equilibrium Kc must increase.

B. False .

3. In order to reach equilibrium CO must be consumed.

A. True.

4. Qc is greater than Kc.

A. True

5. The reaction is at equilibrium. No further reaction will occur.

B. False.

Explanation:

Based on the reaction:

COCl₂(g) → CO (g) + Cl₂(g)

And Kc is defined as:

Kc = 1.29x10⁻² = [CO] [Cl₂] / [COCl₂]

Molar concentrations of each species are:

[COCl₂] = 0.104 moles of COCl₂ / 1L = 0.104M

[CO] = 4.66×10⁻² moles of CO / 1L = 4.66×10⁻²M

[Cl₂] = 3.76×10⁻² moles of Cl₂ / 1L = 3.76×10⁻²M

Replacing in Kc formula:

4.66×10⁻²M × 3.76×10⁻²M / 0.104M = 1.68x10⁻²

As the concentrations are not in equilibrium, 1.68x10⁻² is defined as the <em>reaction quotient, Qc</em>.

As Qc > Kc, the reaction will shift to the left producing more COCl₂ and consuming CO and Cl₂. Thus

1. In order to reach equilibrium COCl₂(g) must be consumed.

B. False

2. In order to reach equilibrium Kc must increase.

B. False . Kc is a constant that never change.

3. In order to reach equilibrium CO must be consumed.

A. True.

4. Qc is greater than Kc.

A. True

5. The reaction is at equilibrium. No further reaction will occur.

B. False. The reaction is in equilibrium when Qc = Kc

6 0
4 years ago
Combustion analysis of 1.200 g of an unknown compound containing carbon, hydrogen, and oxygen produced 2.086 g of CO2 and 1.134
timurjin [86]
<span>the empirical formula is C3H8O2 You need to determine the relative number of moles of hydrogen and carbon. So you first calculate the molar mass of CO2 and H20 Atomic weight of carbon = 12.0107 Atomic weight of hydrogen = 1.00794 Atomic weight of oxygen = 15.999 Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 Molar mass H2O = 2 * 1.00794 + 15.999 = 18.01488 Now calculate the number of moles of CO2 and H2O you have Moles CO2 = 2.086 g / 44.0087 g/mole = 0.0474 mole Moles H2O = 1.134 g / 18.01488 g/mole = 0.062948 mole Calculate the number of moles of carbon and hydrogen you have. Since there's 1 carbon atom per CO2 molecule, the number of moles of carbon is the same as the number of moles of CO2. But since there's 2 hydrogen atoms per molecule of H2O, The number of moles of hydrogen is double the number of moles of H2O Moles Carbon = 0.0474 Moles Hydrogen = 0.062948 * 2 = 0.125896 Now we need to determine how much oxygen is in the compound. Just take the mass of the compound and subtract the mass of carbon and hydrogen. What's left will be the mass of oxygen. Then divide that mass by the atomic weight of oxygen to get the number of moles of oxygen we have. 1.200 - 0.0474 * 12.0107 - 0.125896 * 1.00794 = 0.503797 Moles oxygen = 0.503797 / 15.999 = 0.031489 So now we have a ratio of carbon:hydrogen:oxygen of 0.0474 : 0.125896 : 0.031489 We need to find a ratio of small integers that's close to that ratio. Start by dividing everything by 0.031489 (selected because it's the smallest value) getting 1.505288 : 3.998095 : 1 The 1 for oxygen and the 3.998095 for hydrogen look close enough. But the 1.505288 for carbon doesn't work. But it looks like if we double all the numbers, we'll get something close to an integer for everything. So do so. 3.010575 : 7.996189 : 2 Now this looks good. Rounding everything to an integer gives us 3 : 8 : 2 So the empirical formula is C3H8O2</span>
5 0
3 years ago
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