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Y_Kistochka [10]
3 years ago
11

Which organism obtains energy without depending on another organism

Chemistry
1 answer:
DanielleElmas [232]3 years ago
4 0

Answer:

Plants

Explanation:

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Equivalent massof sodium​
andre [41]

Answer:

Sodium

Formula : Na

Equivalent mass:23.0

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8 0
3 years ago
What is the mass of CO2 lost at 20 min from the limestone sample?
harkovskaia [24]

Answer:

The mass, CO2 and CO3 from the limestone sample is discussed below in details.

Explanation:

(A) mass loss of sample of limestone after 20 min

= 0.8437g-0.5979g = 0.2458 g

From the given reaction of limestone, 2 mol of the sample gives 2 moles of CO 2.

Therefore  

184.4 g ( molar mass of limestone) gives2× 44 g of carbon dioxide.

1 g of sample gives 88/184.4 g of carbon dioxide

Hence 0.2458 g sample gives

= 88/184.4 × 0.2458 g = 0.117 g carbon dioxide

(B) mole of CO 2 lost = weight/ molar mass

= 0.117 g / 44 g/mol =0.0027 mole

(C). 1 mol of limestone contain 2 mol of carbonate ion

From the reaction we know that carbonate ion of limestone is converted into carbondioxide

Hence lost carbonate ion = 0.2458 g

(D) we know that

1 mol limestone contain 1mol CaCO​​​ 3

Hence in sample present CaCO​​​ 3

= 1mole / 184.4 g × 0.8437 g= 0.00458 mol CaCO​​​3

8 0
3 years ago
Hello! could anyone help me? i don't really understand. Thank you!! ​
liq [111]

I'm curious on what it is but I wish I could help

7 0
2 years ago
Determine the oxidation state of each of the following species.<br> Pb2+
mafiozo [28]

Answer:

Pb oxidation number is +2

C in CH4 is -4. H is +1. 4H + 1C = 0. ; 4(+1) + C = 0 ; C = -4

O is usually -2. 3O = 3(-2) = -6. ; 2Fe =. +6 ; each Fe is +3

2Ag = +2 since O = -2 ; each Ag = +1

6 0
3 years ago
How long does it take for a 12.62g sample of ammonia to heat from 209K to 367K if heated at a constant rate of 6.0kj/min? The me
Georgia [21]
First, consider the steps to heat the sample from 209 K to 367K.

1) Heating in liquid state from 209 K to 239.82 K

2) Vaporaizing at 239.82 K

3) Heating in gaseous state from 239.82 K to 367 K.


Second, calculate the amount of heat required for each step.

1) Liquid heating

Ammonia = NH3 => molar mass = 14.0 g/mol + 3*1g/mol = 17g/mol

=> number of moles = 12.62 g / 17 g/mol = 0.742 mol

Heat1 = #moles * heat capacity * ΔT

Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J

2) Vaporization

Heat2 = # moles * H vap

Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J

3) Vapor heating

Heat3 = #moles * heat capacity * ΔT

Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J

Third, add up the heats for every steps:

Total heat = 1,847.77 J + 17,310 J + 3,308.53 J = 22,466.3 J

Fourth, divide the total heat by the heat rate:

Time = 22,466.3 J / (6000.0 J/min) = 3.7 min

Answer: 3.7 min


3 0
3 years ago
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