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3 years ago

When you mix copper sulphate solution and steel wool, what is the chemical property that can be observed.

Chemistry

1 answer:

bazaltina [42]3 years ago

6 0

Answer:

Explanation:

depending on the activity series there will probably be a single replacement reaction possibly heat or color change and the copper precipitate out of solution

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What is the speed of a wave with a wavelength of 3 m and a frequency of .1 Hz?

ratelena [41]

Answer:

wave velocity= frequency × wave length

=1×3

=3m/s

Explanation:

The distance covered by the wave in one second is equal to its wavelength, therefore,

wave velocity=wavelength/time period

OR wave velocity= frequency× wavelength

You can assume velocity as speed here.

3 0

3 years ago

Does the mass number of an atom change when the atom forms a chemical bond?

cupoosta [38]

No it stays the same the only thing that changes are the number of electrons

6 0

3 years ago

(will give brainliest) show your work. How many grams of Copper(I) nitrate, CuNO3 are required to produce 88.0 grams of aluminum

ValentinkaMS [17]

Based on the stoichiometry of the reaction, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

<h3>What is stoichiometry of a reaction?</h3>

The stoichiometry of a reaction is the molar ratio in which reactants combine to form products.

The stoichiometry of the reaction shows that 6 moles of copper (i) nitrate produces 2 moles of aluminium nitrate.

molar mass of Copper(I) nitrate, CuNO3 = 126 g

molar mass of aluminum nitrate, Al(NO3)3 = 213 g

88.0 g of aluminum nitrate, Al(NO3)3 = 88.0/213 moles = 0.413 moles

0.413 moles of Al(NO3)3 will be produced by 0.413 ×6/3 = 1.239 moles of CuNO3

Mass of 1.239 moles of CuNO3 = 1.239 × 126 = 156.114 g of CuNO3

Therefore, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

Learn more about stoichiometry at: brainly.com/question/16060223

Therefore, 156.114 g of CuNO3

4 0

2 years ago

A water sample is found to contain 59 ppm

NeX [460]

Explanation:

k so basically u gotta do 59/1000000 then multiply that by 972 which gives u 0.057348

8 0

4 years ago

A mixture containing 0.477 mol he(g), 0.265 mol ne(g), and 0.115 mol ar(g) is confined in a 7.00-l vessel at 25 ∘c. part a calcu

enot [183]

Q1)
we can use the ideal gas law equation to find the total pressure of the system ;
PV = nRT
where P - pressure
V - volume - 7 x 10⁻³ m³
n - number of moles
total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in K - 273 + 25 °C = 298 K
substituting the values in the equation
P x 7 x 10⁻³ m³ = 0.857 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 303.33 kPa
1 atm = 101.325 kPa
Therefore total pressure - 303.33 kPa / 101.325 kPa/atm = 2.99 atm

Q2)
partial pressure is the pressure exerted by the individual gases in the mixture.
partial pressure for each gas can be calculated by multiplying the total pressure by mole fraction of the individual gas.

total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
mole fraction of He - $\frac{0.477}{0.857} = 0.557$
mole fraction of Ne - $\frac{0.265}{0.857} = 0.309$
mole fraction of Ar - $\frac{0.115}{0.857} = 0.134$
partial pressure - total pressure x mole fraction
partial pressure of He - 2.99 atm x 0.557 = 1.67 atm
partial pressure of Ne - 2.99 atm x 0.309 = 0.924 atm
partial pressure of Ar - 2.99 atm x 0.134 = 0.401 atm

6 0

3 years ago