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nexus9112 [7]
3 years ago
9

Group 1 includes the alkali metals. what are the general properties of elements in this group

Chemistry
1 answer:
pshichka [43]3 years ago
7 0
The general properties of Group 1 alkali metals are:
1. They form cations. 
2. They are highly reactive. 
3. They have a charge of +1.
4. They form ionic compounds. 
5. When their oxides or hydroxides are reacted in water, they form alkalis i.e. bases. 
6. They are usually soft in nature. 
7. They have low densities. 
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4. How many moles of KBr are found in 3 Liters of 0.4 M solution?
Allisa [31]

There are 1.2 moles of KBr found in 3 Liters of 0.4 M solution.

<h3>HOW TO CALCULATE NUMBER OF MOLES?</h3>

The number of moles of a substance can be calculated by multiplying the molarity by the volume.

No. of moles = Molarity × volume

According to this question, 3L of a KBr solution are contained in a 0.4M.

no. of moles = 3L × 0.4M = 1.2moles

Therefore, there are 1.2 moles of KBr found in 3 Liters of 0.4 M solution.

Learn more about no. of moles at: brainly.com/question/14919968

4 0
2 years ago
If the pressure changes from 1 atm to 3 atm, what does the volume of 30 L change to
DiKsa [7]

Answer: 10L

Explanation:

Given that:

Initial pressure P1 = 1 atm

New pressure P2 = 3 atm

Initial volume V2 = 30 L

New volume V2 = ?

Since pressure and volume are involved, apply the formula for Boyle's law

P1V1 = P2V2

1 atm x 30L = 3 atm x V2

30 atm L = 3 atm x V2

V2 = (30 atm L / 3 atm)

V2 = 10L

Thus, volume changed to 10 liters

5 0
3 years ago
A gymnast jumps from a great height onto a sealed mat filled with air.
ololo11 [35]

The gas particles squeeze closer together

6 0
3 years ago
Read 2 more answers
Draw the molecules listed below and circle any functional groups present. Provide the following information for each molecule:Mo
aksik [14]

The question requires us to draw the structural formula, provide the name and highlight any functional groups for the compound: diethyl ether.

The molecule diethyl ether can be represented as it follows, with two ethyl groups (-CH2CH3) bonded to a oxygen atom:

Note that the functional group ether (R-O-R) is present in the structre and highlighted in blue in the image. The official name of diethyl ether is ethoxyethane.

3 0
1 year ago
Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of
bagirrra123 [75]

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

<u>Step 1: </u>Data given

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

<u>Step 2:</u> The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

<u>Step 3:</u> Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

<u>Step 4:</u> Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

<u>Step 5: </u>Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

<u>Step 6:</u> Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

<u>Step 7</u>: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

<u>Step 8:</u> Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

4 0
3 years ago
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