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e-lub [12.9K]
3 years ago
12

How many moles are present in 2.95 x1024 molecules of water (H20)

Chemistry
1 answer:
DerKrebs [107]3 years ago
4 0

ANSWER: 3020.8 moleclues of water (h20)           Explanation I did the math and the quiz

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a balloon is filled with 3.60 L of H2 gas at STP. If the balloon is taken into the sea where the pressure is 2.50 atm and the te
kherson [118]

The new volume of the hydrogen gas in the balloon is 1.5 L.

<u>Explanation:</u>

At STP, Temperature, T1 is 0° C = 0 + 273 K = 273 K

Pressure, P1 = 1 atm

Here given volume, V1 = 3.60 L

Balloon at sea,

Pressure, P2 = 2.5 atm

Temperature, T2 = 10° C = 10 + 273 = 283 K

Volume, V2 = ?

Here, we have to use the equation,

$\frac{P1V1}{T1} = \frac{P2V2}{T2}

We have to rearrange the equation to get V2 as,

$V2 = \frac{P1V1T2}{T1P2}

Now plugin the values as,

V2 = $\frac{1 atm \times 3.6 L \times 283 K}{273 K \times 2.5}

   = 1.5 L

So the new volume of the hydrogen gas is 1.5 L.

7 0
3 years ago
Read 2 more answers
How many grams are in 4.23E44 particles of PbO2
ASHA 777 [7]

1.679 × 10²³ g of PbO₂ is present in 4.23×10⁴⁴ particles of PbO₂.

<u>Explanation:</u>

First the number of particles is converted into moles by dividing it by the Avogadro's number and then moles multiplied by the molar mass, we will get the mass of PbO₂ in grams.

4.23×10⁴⁴/ 6.022×10²³ = 7.02×10²⁰ moles

Now we have to multiply this by the molar mass of PbO₂, that is 239.2 g/mol, we will get the mass in grams.

7.02×10²⁰ moles × 239.2 g/mol = 1.679 × 10²³ g of PbO₂

3 0
3 years ago
Which indicator most likely suggests that a chemical change is taking place?
Nezavi [6.7K]
Chemical change is most often indicated by a change in color.
5 0
4 years ago
Read 2 more answers
If 6.0g of carbon is heated in air the mass of the product obtained could be either 22.0g or 14.0g depending on the amount of ai
Gnom [1K]

In accordance with Dalton's Law of multiple proportions

<h3>Further explanation</h3>

Given

6.0g of carbon

22.0g or 14.0g of product

Required

related laws

Solution

the amount of air present ⇒ as an excess or limiting reactant

  • air(O₂) as a limiting reactant(product=14 g)

C+0.5O₂⇒CO

6 + 8 = 14 g

mol O₂=8 g : 32 g/mol=0.25

mol C = 6 g : 12 g/mol = 0.5(2 x mol O₂)

mol CO= 2 x mol O₂ = 0.5 mol = 0.5 x 28 g/mol = 14 g

  • air(O₂) as an excess reactant(product=22 g) an C as a limiting reactant

C+O₂⇒CO₂

6 + 16 = 22 g

mol C = 6 g : 12 g/mol = 0.5

mol O₂ = 16 g : 32 g/mol=0.5

mol CO₂ = 22 g : 44 g/mol = 0.5

if the mass firs element (C) constant, then the mass of the second element(O) in the two compounds will have a ratio as a simple integer.

CO = 6 : 8

CO₂ = 6 : 16

the ratio O = 8 : 16 = 1 : 2

In accordance with Dalton's Law of multiple proportions

4 0
3 years ago
Which correctly lists the three land uses that the Bureau of Land Management was originally created to manage?
MArishka [77]

Answer:

C. grazing, mining, developing oil and gas.

Explanation:

The Bureau of Land Management's mission is to sustain the health, diversity, and productivity of public lands for the use and enjoyment of present and future generations.

Hope it helps.

7 0
3 years ago
Read 2 more answers
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