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3 years ago

What name should be used for the ionic compound Cu(NO3)2

Chemistry

2 answers:

Sergeu [11.5K]3 years ago

6 0

<span>Copper(II) nitrate. Hope i cleared your doubt</span>

devlian [24]3 years ago

4 0

Answer: The name is Copper(II) nitrate.

Explanation: The name of the ionic compounds is written by writing the name of the cation first followed by its oxidation state in round brackets and then the name of the anion is written without any suffix.

Thus the cation copper is written first as calcium followed by the oxidation state as (II) and then the anion which is nitrate.

Thus the name is Copper(II) nitrate.

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For a given reaction, ΔS = +106 J/mol ⋅ K, and the reaction is spontaneous at temperatures above the crossover temperature, 446

solniwko [45]

Answer: 47.276 kJ/mol

Explanation:

According to Gibb's equation:

$\Delta G=\Delta H-T\Delta S$

$\Delta G$ = Gibbs free energy

$\Delta H$ = enthalpy change

$\Delta S$ = entropy change = +106 J/mol

T = temperature in Kelvin = 446 K

$\Delta G$ = +ve, reaction is non spontaneous

$\Delta G$ = -ve, reaction is spontaneous

$\Delta G$ = 0, reaction is in equilibrium

$\Delta H=T\Delta S$

$\Delta H=446K\times +106J/Kmol=47276J/mol=47.276kJ/mol$ (1kJ=1000J)

Thus the value of ΔH = 47.276 kJ/mol, assuming that ΔH and ΔS do not vary with temperature.

7 0

3 years ago

I need help with this one

Oxana [17]

Answer:

Claim 1 is the correct answer.

8 0

3 years ago

Read 2 more answers

HELP.

Kamila [148]

Answer:

The mass of tin is 164 grams

Explanation:

Step 1: Data given

Specific heat heat of tin = 0.222 J/g°C

The initial temeprature of tin = 80.0 °C

Mass of water = 100.0 grams

The specific heat of water = 4.184 J/g°C

Initial temperature = 30.0 °C

The final temperature = 34.0 °C

Step 2: Calculate the mass of tin

Heat lost = heat gained

Qlost = -Qgained

Qtin = -Qwater

Q = m*c*ΔT

m(tin)*c(tin)*ΔT(tin) = -m(water)*c(water)*ΔT(water)

⇒with m(tin) = the mass of tin = TO BE DETERMINED

⇒with c(tin) = the specific heat of tin = 0.222J/g°C

⇒with ΔT(tin) = the change of temperature of tin = T2 - T1 = 34.0°C - 80.0°C = -46.0°C

⇒with m(water) = the mass of water = 100.0 grams

⇒with c(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = 34.0° C - 30.0 °C = 4.0 °C

m(tin) * 0.222 J/g°C * -46.0 °C = -100.0g* 4.184 J/g°C * 4.0 °C

m(tin) = 163.9 grams ≈ 164 grams

The mass of tin is 164 grams

4 0

3 years ago

The energy of any one-electron species in its nth state (n = principal quantum number) is given by E = –BZ2 /n2 where Z is the c

Ivahew [28]

Explanation:

(a) The given data is as follows.

B = $2.180 \times 10^{-18} J$

Z = 4 for Be

Now, for the first excited state $n_{f}$ = 2; and $n_{i} = \infinity$ if it is ionized.

Therefore, ionization energy will be calculated as follows.

I.E = $\frac{-Bz^{2}}{\infinity^{2}} - (\frac{-2.180 \times 10^{-18} J /times (4)^{2}}{(2)^{2}})$

= $8.72 \times 10^{-18} J$

Converting this energy into kJ/mol as follows.

$8.72 \times 10^{-18} J \times 6.02 \times 10^{23} mol$

= 5249 kJ/mol

Therefore, the ionization energy of the $Be^{3+}$ ion in its first excited state in kilojoules per mole is 5249 kJ/mol.

(b) Change in ionization energy is as follows.

$\Delta E = -Bz^{2}(\frac{1}{(4)^{2}} - {1}{(2)^{2}}) = \frac{hc}{\lambda}$

$\frac{hc}{\lambda} = 0.1875 \times 2.180 \times 10^{-18} J \times (4)^{2}$

$\lambda = \frac{6.626 \times 10^{-34} \times 2.998 \times 10^{8} m/s}{0.1875 \times 2.180 \times 10^{-18} J \times 16}$

= $303.7 \times 10^{-10} m$

or, = $303.7^{o}A$

Therefore, wavelength of light given off from the $Be^{3+}$ ion by electrons dropping from the fourth (n = 4) to the second (n = 2) energy levels $303.7^{o}A$ .

5 0

3 years ago

If 200.0 g of copper(II) sulfate react with an excess of zinc metal, what is the theoretical yield of copper? 1.253 g 50.72 g 79

Helen [10]

1)we need a balanced equation: CuSO₄ + Zn ---> ZnSO₄ + Cu

2) we need to convert the grams of CuSO₄ to moles using the molar mass.

molar mass CuSO₄= 63.5 + 32.0 + (4 x 16.0)= 160 g/mol

$200.0 g CuSO_4 ( \frac{1 mol}{160 grams} )= 1.25 mol CuSO_4$

3) convert moles of CuSO₄ to moles of Cu

$1.25 mol CuSO_4 ( \frac{1 mol Cu}{1 mol CuSO_4} )= 1.25 mol Cu$

4) convert moles of Cu to grams using it's molar mass.

molar mass Cu= 63.5 g/mol

$1.25 mol (\frac{63.5 grams}{1 mol} )= 79.4 grams Cu$

I did it step-by-step as the explanation but you can do all of this in one step.

$200.0 g CuSO_4 ( \frac{1 mol CuSO_4}{160 g} ) ( \frac{1 mol Cu}{1 mol CuSO_4} ) ( \frac{63.5 grams}{1 mol Cu} )= 79.4 grams Cu$

4 0

4 years ago