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xxTIMURxx [149]
3 years ago
8

The density of gold is 19.3. If I have a nugget with the volume of 10cm3, what is the mass of the cube?

Chemistry
2 answers:
aleksklad [387]3 years ago
8 0
<h3>- - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - </h3>

➷ density = mass/volume

Rearrange for mass

mass = density x volume

Substitute the values:

mass = 19.3 x 10

Solve:

mass = 193g

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

Romashka [77]3 years ago
3 0

0.000193 kilograms  , 0.193  gram  or 193 milligrams

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Convert from moles to particles 6.02x1023 particles 1 mole Question 2. 2.7 moles of lithium Question 3. 1.8 moles of sodium chlo
spayn [35]

Considering the definition of Avogadro's number:

  • the number of molecules of lithium is 1.62621×10²⁴ molecules.
  • the number of molecules of sodium chloride is 1.08414×10²⁴ molecules.

<h3>Definition of Avogadro's number</h3>

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole. Avogadro's number applies to any substance.

<h3>Amount of molecules in this case</h3>

You can apply the following rule of three, considering the Avogadro's number:  If 1 mole of lithium contains 6.023×10²³ molecules, 2.7 moles of lithium contains how many molecules?

amount of molecules of lithium= (6.023×10²³ molecules × 2.7 moles)÷1 mole

<u><em>amount of molecules of lithium=1.62621×10²⁴ molecules</em></u>

Finally, the number of molecules of lithium is 1.62621×10²⁴ molecules.

On the other hand you can apply the following rule of three, considering the Avogadro's number:  If 1 mole of sodium chloride​ contains 6.023×10²³ molecules, 1.8 moles of  sodium chloride​ contains how many molecules?

amount of molecules of sodium chloride= (6.023×10²³ molecules × 1.8 moles)÷1 mole

<u><em>amount of molecules of sodium chloride=1.08414 ×10²⁴ molecules</em></u>

Finally, the number of molecules of sodium chloride is 1.08414×10²⁴ molecules.

Learn more about Avogadro's Number:

brainly.com/question/11907018

brainly.com/question/1445383

brainly.com/question/1528951

#SPJ1

4 0
2 years ago
What volume of a 0.200 M HCI solution is needed to neutralize 25.0 L of a 0.250 M NaOH solution? Follow these steps
Zolol [24]

Answer: A little bit confused can you explain what I have to do

Explanation:

3 0
2 years ago
A 115.0-g sample of oxygen was produced by heating 400.0 g of potassium chlorate.2KClO3 Right arrow. 2KCI + 3O2What is the perce
STALIN [3.7K]

Answer:

73.4% is the percent yield

Explanation:

2KClO₃ →  2KCl  + 3O₂

This is a decomposition reaction, where 2 moles of potassium chlorate decompose to 2 moles of potassium chloride and 3 moles of oxygen.

We determine the moles of salt: 400 g . 1. mol /122.5g= 3.26 moles of KClO₃

In the theoretical yield of the reaction we say:

2 moles of potassium chlorate can produce 3 moles of oxygen

Therefore, 3.26 moles of salt, may produce (3.26 . 3) /2 = 4.89 moles of O₂

The mass of produced oxygen is: 4.89 mol . 32 g /1mol = 156.6g

But, we have produced 115 g. Let's determine the percent yield of reaction

Percent yield = (Produced yield/Theoretical yield) . 100

(115g / 156.6g) . 100 = 73.4 %

5 0
3 years ago
What is the volume of 7.50 x 10^24 molecules of NH3 at STP?
Allushta [10]

27.9L

Explanation:

Given parameters:

Number of molecules = 7.5 x 10²⁴molecules

Condition = STP

Unknown:

Volume of gas = ?

Solution;

The volume of gas at STP is expressed below;

 Volume of gas = number of moles x 22.4

Number of moles = \frac{Number of molecules }{6.02 x 10^{23} }

    Now if we can obtain the value of the number of moles from the expression above, we can plug it back into the equation of the volume of gas at STP;

     Number of moles of NH₃ =  \frac{[tex]7.5 x 10^{24} }{6.02 x 10^{23} }[/tex]

   Number of moles of NH₃ = 12.5moles

Volume of gas = 12.5 x 22.4 = 27.9L

learn more:

Volume of gas at STP brainly.com/question/7795301

#learnwithBrainly

3 0
3 years ago
What mass of ammonia can be produced if 13.4 grams of nitrogen gas reacted ?
aivan3 [116]

Answer:

If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia

Explanation:

Step 1: Data given

Mass of nitrogen gas (N2) = 13.4 grams

Molar mass of N2 = 28 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate moles of N2

Moles N2 = Mass N2 / molar mass N2

Moles N2 = 13.4 grams / 28.00 g/mol

Moles N2 = 0.479 moles

Step 4: Calculate moles of NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 0.479 moles N2 we'll produce 2*0.479 = 0.958 moles

Step 5: Calculate mass of NH3

Mass of NH3 = moles NH3 * molar mass NH3

Mass NH3 = 0.958 moles * 17.03 g/mol

Mass NH3 = 16.3 grams

If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia

3 0
3 years ago
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