Explanation:
Below is an attachment containing the solution
The volume of 6.40 grams of O₂ gas at STP is 4.48L (option A). Details about volume can be found below.
<h3>How to calculate volume?</h3>
The volume of a gas can be calculated using the following formula:
p = m/v
Where;
- p = density
- m = mass
- v = volume
According to this question, the mass of O₂ gas at STP is 6.40 grams. The density of the gas at STP is 1.43 g/L.
1.43g/L = 6.4g/V
Volume of O2 = 6.4 ÷ 1.43 = 4.48L
Therefore, the volume of 6.40 grams of O₂ gas at STP is 4.48L.
Learn more about volume at: brainly.com/question/1578538
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<em>A cylinder of argon gas contains 50.0 L of Ar at 18.4 atm and 127 °C. How many moles of argon are in the cylinder?</em>
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Number moles of Argon : 28.03
<h3>Further explanation</h3>
Given
The volume of gas=50 L
P = 18.4 atm
T = 127+273=400 K
Required
moles of Argon
Solution
Use ideal gas Law :
Answer:
Not really fun facts about hazardous waste
Explanation:
four characteristics of hazardous waste are: ignitability • corrosivity • reactivity • toxicity. These things should be thrown away a proper way to reduce pollution.
With that informatio you can:
1) Write the chemical equation
2) Balance the chemical equation
3) State the molar ratios
4) Predict if precipitation occurs.
I will do all four, for you:
1) Chemical equation:
mercury(I) nitrate potassium bromide mercury(I) bromide potassium nitrate
<span>Hg2(NO3)2 + KBr → Hg2Br2 + KNO<span>3
2) Balanced chemical equation
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<span>Hg2(NO3)2 + 2KBr → Hg2Br2 + 2KNO<span>3
3) Molar ratios or proportions:
1 mol </span></span><span>Hg2(NO3)2 : 2 mol KBr : 1 mol Hg2Br2 : 2 mol KNO<span>3
4) Prediction of precipitation.
You can use the solubility rules or a table of solubilities. I found in a table of solutiblities that mercury(I) bromide is insoluble and potassium bromide is soluble, Then you can predict that the precipitation of mercury(I) bromide will occur.
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