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1 year ago

If element x forms the oxides xo and x2o3 the oxidation numbers of element x are

Chemistry

1 answer:

Deffense [45]1 year ago

3 0

Answer:

b) +2 and +3.

Explanation:

Hello,

In this case, given the molecular formulas:

$XO$

And:

$X_2O_3$

We can relate the subscripts with the oxidation states by knowing that they are crossed when the compound is formed, for that reason, we notice that oxygen oxidation state should be -2 for both cases and the oxidation state of X in the first formula must be +2 since both X and O has one as their subscript as they were simplified:

$X^{+2}O^{-2}$

Moreover, for the second case the oxidation state of X should be +3 in order to obtain 3 as the subscript of oxygen:

$X_2^{+3}O_3^{-2}$

Thus, answer is b)+2 and +3

Best regards.

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A student has 7.05 g of zinc powder, 1.60 L of a 3.40 M calcium nitrate solution, and 1.50 L of a 1.60 M lead(II) nitrate soluti

trapecia [35]

Answer: Zinc will react with lead (II) nitrate solution.

Explanation:

Single displacement reaction is defined as the reaction in which more reactive element displaces a less reactive element from its chemical reaction.

The reactivity of metal is determined by a series known as reactivity series. The metals lying above in the series are more reactive than the metals which lie below in the series.

General equation for single displacement reaction follows:

$A+BC\rightarrow AC+B$

When zinc is reacted with calcium nitrate, the reaction does not take place as zinc is less reactive than calcium. Zinc lies below in the series than calcium.

$Zn+Ca(NO_3)_2\rightarrow \text{No reaction}$

But, when zinc is reacted with lead (II) nitrate, the reaction do take place as zinc is more reactive than lead. Zinc lies above in the series than lead.

The chemical equation for the reaction of zinc and lead (II) nitrate follows:

$Zn+Pb(NO_3)_2\rightarrow Zn(NO_3)_2+Pb$

Hence, zinc will react with lead (II) nitrate solution.

6 0

1 year ago

Convert 1.71 × 1024 atoms of carbon to moles of carbon.

beks73 [17]

You need to use Avogadro's constant to convert from atoms of carbon to moles of carbon.

1.71*10^24 atoms C * (1 mole/6.022*10^23 atoms) = 2.84 moles of carbon

7 0

2 years ago

Calculate the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 3

ZanzabumX [31]

Answer:

19,26 kJ

Explanation:

The work done when a gas expand with a constant atmospheric pressure is:

W = PΔV

Where P is pressure and ΔV is the change in volume of gas.

Assuming the initial volume is 0, the reaction of 500g of Zn with H⁺ (Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)) produce:

500,0g Zn(s)× $\frac{1molZn}{65,38g}$ × $\frac{1molH_{2}(g)}{1molZn}$ = 7,648 moles of H₂

At 1,00atm and 303,15K (30°C), the volume of these moles of gas is:

V = nRT/P

V = 7,648mol×0,082atmL/molK×303,15K / 1,00atm

V = 190,1L

That means that ΔV is:

190,1L - 0L = 190,1L

And the work done is:

W = 1atm×190,1L = 190,1atmL.

In joules:

190,1 atmL× $\frac{101,325}{1atmL}$ = 19,26 kJ

I hope it helps!

5 0

2 years ago

Relate dark matter to the development of the universe after the Big Bang. In 3-5 sentences, speculate on how the development of

Alecsey [184]

Answer:

Without dark matter galaxies would loose an extreme amount of gas required to create stars.

Without dark matter the universe wont have as many galaxies clumped together forming larger versions of those galaxies. This would cause a change in the structure of the "skeleton" of the web.

(Hope this can help, I didn't do exactly as it is said to because that is your job)

Explanation:

Forbes gives somewhat of an explanation if you are curious.

(Ethan Siegal, "The Universe Would Be Very Different Without Dark Matter", Forbes)

3 0

1 year ago

A 5.00 L sample of helium expands to 12.0 L at which point the

mina [271]

Answer:

1.73 atm

Explanation:

Given data:

Initial volume of helium = 5.00 L

Final volume of helium = 12.0 L

Final pressure = 0.720 atm

Initial pressure = ?

Solution:

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume

Now we will put the values in formula,

P₁V₁ = P₂V₂

P₁ × 5.00 L = 0.720 atm × 12.0 L

P₁ = 8.64 atm. L/5 L

P₁ = 1.73 atm

7 0

1 year ago