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12345 [234]
3 years ago
6

How many mL of water would be displaced by 408 g of lead?

Chemistry
1 answer:
Galina-37 [17]3 years ago
4 0

Answer:

36 mL of water can be displace by the 408 g lead

Explanation:

Given data:

Mass of lead = 408 g

Volume of water displaced by lead = ?

Solution:

The volume of water displace by the lead would be equal to the given volume of lead.

So we will calculate the volume from given mass of lead from density formula.

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

Density of lead is 11.36 g/cm³.

d = m/v

11.36 g/cm³ = 408 g/ v

v = 408 g/ 11.36 g/cm³

v =36 cm³

1 cm³  = 1 mL

volume = 36 mL

36 mL of water can be displace by the 408 g lead.

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Nookie1986 [14]

Answer:

643g of methane will there be in the room

Explanation:

To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -<em>Assuming STP-</em>. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:

<em>Volume Methane:</em>

3600s * (0.25L / s) = 900L Methane

<em>Moles methane:</em>

PV = nRT; PV / RT = n

<em>Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP</em>

Replacing:

PV / RT = n

1atm*900L / 0.082atmL/molK*273.15 = n

n = 40.18mol methane

<em>Mass methane:</em>

40.18 moles * (16g/mol) =

<h3>643g of methane will there be in the room</h3>
5 0
3 years ago
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B) 40%  
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4) How many moles of O2 are produced with 42.0 moles of H2O?
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Explanation:

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<h2>Answer:</h2>

The correct answer is option D which is substances will pass in and out of the cell in an uncontrolled manner.

<h3>Explanation:</h3>
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