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DENIUS [597]
3 years ago
15

An average quality lawn captures ______ times the carbon output of a typical gasoline-powered mower

Chemistry
1 answer:
valina [46]3 years ago
5 0
An average quality lawn captures 4 times the carbon output of a typical gasoline-powered mower.
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Beth’s hygrometer is reading a temperature of 30°C and a relative humidity of 65%. The humidity in the air is how many grams/cub
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This is asking for the absolute humidity. So if the relative humidity is 65 % and the air temperature is 30 then we need to have in mind the barometric pressure which will be 760. The humidity in the air will be 0.020 kg/m3. In order to know this we calculate it eith the mass of water vapour in a unit volume of air. This is a measure of the actual water vapour content of the air. the for mula would go like this: AH = Mv / V. 
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2 years ago
20 points and brainliest for best answer
RSB [31]
Gold/Atomic mass

196.96657 u ± 0.000004 u

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39.0983 u ± 0.0001 u

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Which is the best analogy of a wave?
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How many moles of oxygen are formed when 58.6 g of KNO3 decomposes according to the following reaction? 4 KNO3(s) → 2 K2O(s) + 2
Leya [2.2K]

Answer:

0.725 mol

Explanation:

Moles are calculated as the given mass divided by the molecular mass.

i.e. ,

moles = ( mass / molecular mass )

since,

mass of KNO₃ = 58.6 g  ( given )

Molecular mass of KNO₃ = 101 g / mol

Therefore,

moles of KNO₃ = 58.6 g / 101 g / mol

moles of KNO₃ = 0.58 mol

From the balanced reaction ,

4 KNO₃ (s) ---> 2K₂O (s) + 2N₂ (g) + 5O₂ (g)

By the decomposition of 4 mol of KNO₃ , 5 mol of O₂ are formed ,

hence, unitary method is used as,

1  mol of KNO₃  gives 5 / 4 mol O₂

Therefore,

0.58 mol of KNO₃ , gives , 5 / 4  * 0.58 mol of O₂

Solving,

0.58 mol of KNO₃ , gives , 0.725 mol of O₂

Therefore,

58.6g of KNO₃ gives 0.725 mol of O₂.

3 0
2 years ago
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