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Goryan [66]
3 years ago
11

Chemical formula for tin iv carbide

Chemistry
1 answer:
aniked [119]3 years ago
3 0

Answer:

Explanation:

Sn(WC)2

if it is tungsten carbide this should be correct but there are many versions of carbide

Sn(MC2)2

could also be possible

the 2 next to MC should be a subscript

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Triglycerides, fatty acids and glycerol differ with respect to polarity, H-bonds, and functional groups. Discuss the differences
Black_prince [1.1K]

Answer:

Triglycerides are actually fats made from condensation of fatty acids and glycerol, and used in making soap because it readily reacts caustic alkali and precipitates soap molcules while glycerol is the produced alongside

Explanation:

Triglycerides are made when 3 molecules of fatty acids condenses with one molecule of glycerol having 3-sites of OH where the condensation takes place with the COOH functional group in the fatty acids and 3 molecules of water

4 0
3 years ago
Write word equation for reaction between sodium and sulphur​
White raven [17]

Answer:

2Na + S —> Na2S

Hope this helps.

4 0
3 years ago
What is the redox half equation for 3Ag2S + 2Al --> 6Ag + Al2, and identity which material is oxidized and which is reduced?
marta [7]

Answer:

Al is oxidized while Ag is reduced.

Explanation:

The complete molecular equation is;

3Ag2S + 2Al --> 6Ag + Al2S3

Oxidation half equation;

2Al ------> 2Al^3+ + 6e

Reduction half equation;

6Ag^+ + 6e -------> 6Ag

Overall redox reaction equation;

2Al + 6Ag^+ ----->2Al^3+ + 6Ag

Hence; Al is oxidized while Ag is reduced.

5 0
3 years ago
Two systems with heat capacities 19.9 J mol-1 K-1 and 28.2 ] mol 1 K-1 respectively interact thermally and come to an equilibriu
MAVERICK [17]

Answer : The initial temperature of system 2 is, 19.415^oC

Explanation :

In this problem we assumed that the total energy of the combined systems remains constant.

-q_1=q_2

m\times c_1\times (T_f-T_1)=-m\times c_2\times (T_f-T_2)

The mass remains same.

where,

C_1 = heat capacity of system 1 = 19.9 J/mole.K

C_2 = heat capacity of system 2 = 28.2 J/mole.K

T_f = final temperature of system = 30^oC=273+30=303K

T_1 = initial temperature of system 1 = 45^oC=273+45=318K

T_2 = initial temperature of system 2 = ?

Now put all the given values in the above formula, we get

-19.9J/mole.K\times (303-318)K=28.2J/mole.K\times (303-T_2)K

T_2=292.415K

T_2=292.415-273=19.415^oC

Therefore, the initial temperature of system 2 is, 19.415^oC

8 0
4 years ago
How much energy (in Joules) would it take to warm 3.11 grams of gold by 7.7 oC
Reptile [31]

Answer:

It would take 3.11 J to warm 3.11 grams of gold

Explanation:

Step 1: Data given

Mass of gold = 3.11 grams

Temperature rise = 7.7 °C

Specific heat capacity of gold = 0.130 J/g°C

Step 2: Calculate the amount of energy

Q = m*c*ΔT

⇒ Q = the energy required (in Joules) = TO BE DETERMINED

⇒ m = the mass of gold = 3.11 grams

⇒ c = the specific heat of gold = 0.130 J/g°C

⇒ ΔT = The temperature rise = 7.7 °C

Q = 3.11 g * 0.130 J/g°C * 7.7 °C

Q = 3.11 J

It would take 3.11 J to warm 3.11 grams of gold

4 0
3 years ago
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