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EleoNora [17]
3 years ago
8

How many grams are in 0.02 moles of beryllium iodide

Chemistry
1 answer:
saul85 [17]3 years ago
5 0
Molar mass beryllium iodide ( BeI₂ ) = <span>262.821 g/mol

mass = number of moles * molar mass

mass = 0.02 * 262.821

mass ( BeI</span>₂) = 5.25642 g
<span>
hope this helps!

</span>
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A solution is prepared by dissolving sugar in water. The solution is 25% by mass, sugar. How many grams of water are in 472 gram
stepan [7]

Answer:

352

Explanation:

.75 times 472 because .25 is sugar so .75 is watet

4 0
3 years ago
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Which of the following is the correct model of C6H14?
asambeis [7]

Answer:

This question is incomplete

Explanation:

This question is incomplete because of the absence of options. However, the compound C₆H₁₄ is hexane. Hexane is a member of saturated hydrocarbons (homologous series) called alkanes (with the general formula CₙH₂ₙ₊₂). The structure for an hexane is shown below

     H   H   H   H   H   H

      I     I    I     I     I     I

H - C - C - C - C - C - C - H

      I     I     I    I      I    I

     H   H   H   H    H  H

which can also be written as

CH₃CH₂CH₂CH₂CH₂CH₃

3 0
3 years ago
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the density of aluminum is 2.70g/cm^3. a piece of aluminum foil has a volume of 54.0 cm^3. what is the mass of this piece of alu
Neporo4naja [7]
The  mass  of  aluminium  foil   is   calculated   as  follows
mass  =  density  x  volume
density =  2.70  g/cm^3
volume  54  cm^3
mass  of   aluminium  foil  is  therefore  =  2.70  g/cm^3  x  54  cm^3  =145.8  grams
cm^3  cancel   out  each   other
6 0
3 years ago
A gas at 29.4 kPa is cooled from a temperature of 75°C to a temperature of 25°C at constant volume. What is the new pressure of
kirill115 [55]

<span>To solve this we assume that the gas inside the balloon is an ideal </span>gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant volume pressure and number of moles of the gas the ratio of T and P is equal to some constant. At another set of condition, the constant is still the same. Calculations are as follows:

 

T1/P1 = T2/P2

P2 = T2 x P1 / T1

P2 = 25 x 29.4 / 75

P2 = 9.8 kPa

7 0
3 years ago
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What is the percent yield if 107.50 g NH3 reacts with excess O2 according to the
Maksim231197 [3]

Answer:

81.59%

Explanation:

  • 4NH₃ + 5O₂ → 4NO + 6H₂O

First we <u>convert 107.50 g of NH₃ into moles</u>, using its <em>molar mass</em>:

  • 107.50 g NH₃ ÷ 17 g/mol = 6.32 mol NH₃

Now we <u>calculate how many moles of NO would have been formed by the complete reaction of 6.32 moles of NH₃</u>:

  • 6.32 mol NH₃ * \frac{4molNO}{4molNH_3} = 6.32 mol NO

Then we <u>convert 6.32 moles of NO to grams</u>, using its <em>molar mass</em>:

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Finally we <u>calculate the percent yield</u>:

  • 154.70 g / 189.60 g * 100% = 81.59%

8 0
3 years ago
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