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nadya68 [22]
3 years ago
5

How do you convert 0.85 atm to mmHg

Chemistry
1 answer:
laiz [17]3 years ago
8 0

1 atm = 760 mm Hg

so, (0.85atm)*(760mmHg/1atm) = 646mmHg


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Diatomic gases such as H2(g), O2(g), and N2(g) contain ________ bonds.
Studentka2010 [4]
Diatomic gases contain covalent bond
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3 years ago
Discuss the reason for the presence of large number of organic compounds
jeyben [28]

Answer:

Carbon forms the large numbers of compound due to the following reasons

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3 years ago
Which kind of plate boundary created the Mid-Atlantic Ridge or the chain of volcanoes underneath Atlantic Ocean?
Natali [406]

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Divergent plate boundary

Explanation:

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3 years ago
The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) C
Kipish [7]

Answer: 72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Putting in the values we get:

\text{Number of moles}=\frac{19.7g}{44g/mol}=0.45moles

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)

According to stoichiometry:

1 mole of C_3H_8 requires 5 moles of oxygen

0.45 moles of C_3H_8 require= \frac{5}{1}\times 0.45=2.25 moles of oxygen

Mass of O_2=moles\times {\text {Molar mass}}=2.25\times 32=72g

72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

7 0
3 years ago
what volume of co2 is produced at stp when 270g of glucose are consumed in the following reaction? c6h12o6 + 6o2(g) -> 6co2 (
lana [24]

Answer:

202 L

Explanation:

Step 1: Write the balanced equation

C₆H₁₂O₆ + 6 O₂(g) ⇒ 6 CO₂(g) + 6 H₂O(l)

Step 2: Calculate the moles corresponding to 270 g of C₆H₁₂O₆

The molar mass of C₆H₁₂O₆ is 180.16 g/mol.

270 g × 1 mol/180.16 g = 1.50 mol

Step 3: Calculate the moles of CO₂ generated from 1.50 moles of glucose

The molar ratio of C₆H₁₂O₆ to CO₂ is 1:6. The moles of CO₂ formed are 6/1 × 1.50 mol = 9.00 mol

Step 4: Calculate the volume of 9.00 moles of CO₂ at STP

The volume of 1 mole of an ideal gas at STP is 22.4 L.

9.00 mol × 22.4 L/mol = 202 L

4 0
3 years ago
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