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4 years ago

8

2. Atoms are stable and not likely to react when.

1 answer:

Alisiya [41]4 years ago

4 0

4. They aren’t stable

You might be interested in

Why does lithium form only lithium oxide and not peroxide or superoxide

Alexandra [31]

Answer:

Lithium does form a peroxide as well as an oxide on burning in air and I suspect the low temperature reaction with air forms a significant amount of peroxide.

8 0

3 years ago

The names and chemical formulae of some chemical compounds are written in the first two columns of the table below. Each compoun

charle [14.2K]

Answer:

See below

Explanation:

<u> Name </u> <u>Formula </u> <u> Major species </u> <u> </u>

Zinc iodide ZnI₂ H₂O(ℓ), I⁻(aq), Zn²⁺(aq),

Nitrogen(I) oxide N₂O H₂O(ℓ), N₂O(aq)

Sodium nitrite NaNO₂ H₂O(ℓ), Na⁺(aq), NO₂⁻(aq)

Glucose C₆H₁₂O₆ H₂O(ℓ), C₆H₁₂O₆(aq)

Nickel(II) iodide NiI₂ H₂O(ℓ), I⁻(aq), Ni²⁺(aq)

Glucose and nitrogen(I) oxide are covalent compounds. They do not dissociate in solution.
The compounds containing metals are ionic. They produce ions in solution.
ZnI₂ and NiI₂ produce twice as many iodide ions as metal ions.

6 0

3 years ago

The most common source of copper (cu) is the mineral chalcopyrite (cufes2). how many kilograms of chalcopyrite must be mined to

tigry1 [53]

Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.

Solution : Given,

Mass of Cu = 300 g

Molar mass of Cu = 63.546 g/mole

Molar mass of $CuFeS_2$ = 183.511 g/mole

First we have to calculate the moles of Cu.

$\text{ Moles of Cu}=\frac{\text{ Given mass of Cu}}{\text{ Molar mass of Cu}}= \frac{300g}{63.546g/mole}=4.7209moles$

The moles of Cu = 4.7209 moles

From the given chemical formula, $CuFeS_2$ we conclude that the each mole of compound contain one mole of Cu.

So, The moles of Cu = Moles of $CuFeS_2$ = 4.4209 moles

Now we have to calculate the mass of $CuFeS_2$ .

Mass of $CuFeS_2$ = Moles of $CuFeS_2$ × Molar mass of $CuFeS_2$ = 4.4209 moles × 183.511 g/mole = 866.337 g

Mass of $CuFeS_2$ = 866.337 g = 0.8663 Kg (1 Kg = 1000 g)

Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.

3 0

3 years ago

What is △n for the following equation in relating Kc to Kp?

Nimfa-mama [501]

Answer:

-1

Explanation:

The relation between Kp and Kc is given below:

$K_p= K_c\times (RT)^{\Delta n}$

Where,

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant , 0.082057 L atm.mol⁻¹K⁻¹

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

$2Na_{(s)}+2H_2O_{(l)}\rightleftharpoons 2NaOH_{(aq)}+2H_2_{(g)}$

<u>Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants) = (2+1)-(2+2) = -1 </u>

<u></u>

7 0

4 years ago

What is the entropy change in the environment when 5.0 MJ of energy is transferred thermally from a reservoir at 1000 K to one a

Leni [432]

Answer:

The entropy change in the environment is 3.62x10²⁶.

Explanation:

The entropy change can be calculated using the following equation:

$\Delta S = \frac{Q}{k_{B}}(\frac{1}{T_{f}} - \frac{1}{T_{i}})$

Where:

Q: is the energy transferred = 5.0 MJ

$k_{B}$ : is the Boltzmann constant = 1.38x10⁻²³ J/K

$T_{i}$ : is the initial temperature = 1000 K

$T_{f}$ : is the final temperature = 500 K

Hence, the entropy change is:

$\Delta S = \frac{5.0 \cdot 10^{6} J}{1.38 \cdot 10^{-23} J/K}(\frac{1}{500 K} - \frac{1}{1000 K}) = 3.62 \cdot 10^{26}$

Therefore, the entropy change in the environment is 3.62x10²⁶.

I hope it helps you!

7 0

3 years ago