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horsena [70]
3 years ago
14

Question 1

Chemistry
1 answer:
Maksim231197 [3]3 years ago
6 0

Answer:

Hydrogen bonding, Dipole-dipole

Explanation:

Let us look closely at the structure of H2S, this will enable us to decide if it will show dipole-dipole interaction or not. Given that Sulphur has a greater electro negativity compared to Hydrogen, we expect that the S – H bond will be a polar bond with the direction of polarity pointing towards the sulphur atom according to the usual convention in chemistry. Now, we know that H2S is a bent molecule (there are two lone pairs on the central sulphur atom which leads to a bent molecular geometry), therefore, the vectorial sum of the bond dipole moments will produce a non- zero total dipole moment. This implies that there is a permanent non-zero dipole moment in H2S therefore we expect the molecule to exhibit dipole-dipole interactions .

Similarly, we know that hydrogen bonding exists when hydrogen is bonded to a highly electronegative element such as fluorine, sulphur, oxygen etc. In H2S, hydrogen is bonded to sulphur hence we expect that there should be hydrogen bonding between H2S molecules. Hydrogen bonding is actually a type of dipole-dipole interaction.

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What volume of O2 collected at 22.0 and 728 mmHg would be produce by the decomposition of 8.15 g KClO3?
adell [148]

Answer:

There is 2.52 L of O2 collected

Explanation:

Step 1: Data given:

Temperature = 22.0 °C

Pressure = 728 mmHg = 728 /760 = 0.958 atm

Mass of KClO3 = 8.15 grams

Molar mass of KClO3 = 122.55 g/mol

Step 2: The balanced equation

2KClO3(s) → 2KCl(s) + 3O2(g)

Step 3: Calculate moles of KClO3

Moles KClO3 = mass KClO3 / molar mass KClO3

Moles KClO3= 8.15 grams / 122.55 g/mol

Moles KClO3 = 0.0665 moles

Step 4: Calculate moles of O2

For 2 moles of KClO3 we'll have 2 moles of KCl and 3 moles of O2 produced

For 0.0665 moles of KClO3 we have 3/2 * 0.0665 = 0.09975 moles

Step 5: Calculate vlume of O2

p*V = n*R*T

V = (n*R*T)/p

⇒ with n = the number of moles O2 = 0.09975 moles

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

⇒ with T = 22.0 °C = 273 +22 = 295 Kelvin

⇒ with p = 0.958 atm

V = (0.09975 * 0.08206 * 295) / 0.958

V = 2.52 L

There is 2.52 L of O2 collected

7 0
2 years ago
For the following electron-transfer reaction:
creativ13 [48]

Answer:

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)  

Explanation:

Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)

In the oxidation half reaction, the oxidation number increases:

Mn changes from 0, in the ground state to Mn²⁺.

The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.

Silver changes from Ag⁺ to Ag.

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)  

To balance the hole reaction, we need to multiply by 2, the second half reaction:

Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2

2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)  

Now we sum, and we can cancel the electrons:

2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻

4 0
3 years ago
How much heat is required to raise the temperature of 670g of water from 25.7"C to 66,0°C? The specific heat
inna [77]

Answer:

Explanation:

q= mc theta

where,

Q = heat gained

m = mass of the substance = 670g

c = heat capacity of water= 4.1 J/g°C    

theta =Change in temperature=( 66-25.7)

Now put all the given values in the above formula, we get the amount of heat needed.

q= mctheta

q=670*4.1*(66-25.7)

  =670*4.1*40.3

=110704.1

8 0
3 years ago
Water is Carried to the leaves by stomata
tiny-mole [99]

Answer:

please explain further and i maybe can help you

Explanation:

5 0
2 years ago
What is the atomic mass number of an isotope that has an atomic number of 5 and includes 6 neutrons?
irinina [24]

The answer should be; 11

The atomic mass number is found by combining the number of protons and neutrons

Hope this helps :)

8 0
3 years ago
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