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yKpoI14uk [10]
2 years ago
9

How many grams of copper (II) hydroxide can be prepared from 2.4 grams of copper (II) nitrate (Cu(NO3)2 ) and excess sodium hydr

oxide?
Chemistry
1 answer:
andreyandreev [35.5K]2 years ago
5 0

Answer:

percentage yield = 67%

Explanation:

Mass of Cu(NO₃)₂  = 15.25 g

Mass of NaOH   = 12.75 g

Percentage yield = ?

Solution:

Cu(NO₃)₂ + 2NaOH   →  Cu(OH)₂ + 2NaNO₃

Moles of Cu(NO₃)₂:

Number of moles = mass/ molar mass

Number of moles = 15.25 g /187.56 g/mol

Number of moles = 0.08 mol

Moles of NaOH :

Number of moles = mass/ molar mass

Number of moles = 12.75 g / 40 g/mol

Number of moles = 0.32 mol

Now we will compare the moles of Cu(OH)₂ with NaOH and Cu(NO₃)₂.      NaOH             :      Cu(OH)₂

                               2                   :          1

                               0.32              :           1/2×0.32 = 0.16 mol

                            Cu(NO₃)₂         :           Cu(OH)₂

                                  1                  :               1

                             0.08                :              0.08

The number of moles produced by  Cu(NO₃)₂  are less so it will limiting reactant.

Mass of Cu(OH)₂:

Mass = number of moles × molar mass

Mass = 0.08 mol × 97.6 g/mol

Mass = 7.808 g

Theoretical yield = 7.808 g

Percent yield:

percentage yield = Actual yield/ theoretical yield ×  100

percentage yield = 5.23 g/  7.808 g ×  100

percentage yield = 0.67 ×  100

percentage yield = 67%

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An independent variable is the variable changed or controlled in a scientific experiment to test the effects on the dependent variable. A dependent variable is the variable being tested and measured in a scientific experiment.
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2 years ago
Does anyone know what convection,conduction,radiation??
saw5 [17]

Answer:

Conduction is the transfer of thermal energy through direct contact. Convection is the transfer of thermal energy through the movement of a liquid or gas. Radiation is the transfer of thermal energy through thermal emission.

3 0
3 years ago
Part 1. determine the molar mass of a 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm. show your work.
d1i1m1o1n [39]
  • The molar mass of 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm is 9.15g/mol.
  • If this sample was placed under extreme pressure, the volume of the sample will decrease.

<h3>How to calculate molar mass?</h3>

The molar mass of a substance can be calculated by first calculating the number of moles using ideal gas law equation:

PV = nRT

Where;

  • P = pressure
  • V = volume
  • T = temperature
  • R = gas law constant
  • n = no of moles

0.98 × 1.2 = n × 0.0821 × 287

1.18 = 23.56n

n = 1.18/23.56

n = 0.05moles

mole = mass/molar mass

0.05 = 0.458/mm

molar mass = 0.458/0.05

molar mass = 9.15g/mol

  • Therefore, the molar mass of 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm is 9.15g/mol
  • If this sample was placed under extreme pressure, the volume of the sample will decrease.

Learn more about gas law at: brainly.com/question/12667831

8 0
1 year ago
Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the 239Pu present today will be present in 1000 y
olya-2409 [2.1K]

Answer:

0.9715 Fraction of Pu-239 will be remain after 1000 years.

Explanation:

\lambda =\frac{0.693}{t_{\frac{1}{2}}}

A=A_o\times e^{-\lambda t}

Where:

\lambda= decay constant

A_o =concentration left after time t

t_{\frac{1}{2}} = Half life of the sample

Half life of Pu-239 = t_{\frac{1}{2}}=24,000 y[

\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]

Let us say amount present of  Pu-239 today = A_o=x

A = ?

A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}

A=0.9715\times x

\frac{A}{A_0}=\frac{A}{x}=0.9715

0.9715 Fraction of Pu-239 will be remain after 1000 years.

7 0
3 years ago
Calculate the force applied to a car with mass of 1200 kg that has an acceleration of 2 m/s/s.
bearhunter [10]

Answer:

2,400

Explanation:

F = m × a

F = 1200 × 2

F = 2,400

6 0
3 years ago
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