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mel-nik [20]
3 years ago
8

The alkali metals all react with water. which is the most reactive metal?

Chemistry
1 answer:
kifflom [539]3 years ago
8 0
Alkali Metals (Group 1) elements experience an increase in the vigour of their reaction in water as they go down the group (as the atomic number increase). As such the most reactive Alkali Metal would be FRANCIUM, which is at the base of Group One.


Quite frankly, you do not want Francium to react with water- that's a huge explosion on your hand.

You might be interested in
Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t
myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol

Moles of methane gas = n_1=\frac{8.00 g}{16 g/mol}=0.5 mol

Moles of ethane gas  =n_2=\frac{18.0 g}{30 g/mol}=0.6 mol

Moles of propane gas = n_3

n=n_1+n_2+n_3

2.22=0.5 mol +0.6 mol+ n_3

n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol

Mole fraction of methane =\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}

\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252

Similarly, mole fraction of ethane and propane :

\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703

\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045

Partial pressure of each gas can be calculated by the help of Dalton's' law:

p_i=P\times \chi_1

Partial pressure of methane gas:

p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm

Partial pressure of ethane gas:

p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm

Partial pressure of propane gas:

p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = n_1=\frac{37.8 g g}{28g/mol}=1.35 mol

Moles of oxygen gas  =n_2=\frac{62.2 g}{32 g/mol}=1.94 mol

Mole fraction of nitrogen=\chi_1=\frac{n_1}{n_1+n_2}

\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103

Similarly, mole fraction of oxygen

\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897

Partial pressure of each gas can be calculated by the help of Dalton's' law:

p_i=P\times \chi_1

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg

Partial pressure of oxygen gas:

p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg

3 0
3 years ago
PLEASE HELP 20 POINTS What is a key event found in CO2 and glucose?
irina [24]

the answer is carbon

5 0
3 years ago
Refer to the example about diatomic gases A and B in the text to do problems 19 - 27. How many grams in 3 moles of A2?
dedylja [7]

B.) 3

hope you have a okay day

(GAME OF THRONES RULES

7 0
3 years ago
Read 2 more answers
in order to find the molar mass of an unknown compound, a research scientist prepared a solution of 0.930 g of an unknown in 125
PtichkaEL [24]

Answer:

Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

Explanation:

Let's apply the formula for freezing point depression:

ΔT = Kf . m

ΔT = 74.2°C - 73.4°C → 0.8°C

Difference between the freezing T° of pure solvent and freezing T° of solution

Kf = Cryoscopic constant → 5.5°C/m

So, if we replace in the formula

ΔT = Kf . m → ΔT / Kf = m

0.8°C / 5.5 m/°C = m → 0.0516 mol/kg

These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg

0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:

Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

3 0
3 years ago
Balance each of the following redox reactions occurring in basic solution.MnO−4(aq)+Br−(aq)→MnO2(s)+BrO−3(aq)Express your answer
Ahat [919]

Answer : The balanced chemical equation is,

2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)

Explanation :

Rules for the balanced chemical equation in basic solution are :

  • First we have to write into the two half-reactions.
  • Now balance the main atoms in the reaction.
  • Now balance the hydrogen and oxygen atoms on both the sides of the reaction.
  • If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.
  • If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion (OH^-) at that side where the less number of hydrogen are present.
  • Now balance the charge.

The half reactions in the basic solution are :

Reduction : MnO_4^-(aq)+2H_2O(l)+3e^-\rightarrow MnO_2(s)+4OH^-(aq) ......(1)

Oxidation : Br^-(aq)+6OH^-(aq)\rightarrow BrO_3^-(aq)+3H_2O(l)+6e^-  .......(2)

Now multiply the equation (1) by 2 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)

8 0
2 years ago
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