Answer:
An inverse relationship can neither be represented by a straight line nor by a bar chart. But it can be represented by "xy = k"
Explanation:
Inverse relation is used for the values which are inversely related to each other. For example: Let suppose you have a value x and y. Then an increase in the value of x will result in the decrease of value y. Mathematically it is represented as,
x ∝ 1 / y
Where;
∝ = proportionality
Replacing the proportionality sign by a constant value "k" the relation becomes,
x = k / y
Solving for k,
x y = k
Conclusion:
Hence, an inverse relationship can be represented by "xy = k"
There are 0.566 moles of carbonate in sodium carbonate.
<h3>CALCULATE MOLES:</h3>
- The number of moles of carbonate (CO3) in sodium carbonate (Na2CO3) can be calculated by dividing the mass of carbonate in the compound by the molar mass of the compound.
- no. of moles of CO3 = mass of CO3 ÷ molar mass of Na2CO3
- Molar mass of Na2CO3 = 23(2) + 12 + 16(3)
- = 46 + 12 + 48 = 106g/mol
- mass of CO3 = 12 + 48 = 60g
- no. of moles of CO3 = 60/106
- no. of moles of CO3 = 0.566mol
- Therefore, there are 0.566 moles of carbonate in sodium carbonate.
Learn more about number of moles at: brainly.com/question/1542846
Answer:
4
Explanation:
cuz i just took a test and the question was this just reversed. if the ph is 4 and the other ph is 100x greater it’s 6. i don’t kno the reasoning lol
Answer:
The rate at which ammonia is being produced is 0.41 kg/sec.
Explanation:
Haber reaction
Volume of dinitrogen consumed in a second = 505 L
Temperature at which reaction is carried out,T= 172°C = 445.15 K
Pressure at which reaction is carried out, P = 0.88 atm
Let the moles of dinitrogen be n.
Using an Ideal gas equation:


According to reaction , 1 mol of ditnitrogen gas produces 2 moles of ammonia.
Then 12.1597 mol of dinitrogen will produce :
of ammonia
Mass of 24.3194 moles of ammonia =24.3194 mol × 17 g/mol
=413.43 g=0.41343 kg ≈ 0.41 kg
505 L of dinitrogen are consumed in 1 second to produce 0.41 kg of ammonia in 1 second. So the rate at which ammonia is being produced is 0.41 kg/sec.