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4 years ago

7

When electrons are added to an element or compound this is called?

1 answer:

dexar [7]4 years ago

4 0

Answer:

The answer is replacement.

Explanation:

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How does physical weathering happen?

Ilya [14]

Answer:

Physical weathering is caused by the effects of changing temperature on rocks, causing the rock to break apart. The process is sometimes assisted by water. Freeze-thaw occurs when water continually seeps into cracks, freezes and expands, eventually breaking the rock apart

Explanation:

4 0

3 years ago

The diagram shows the electron configuration of an atom of an element for the electrons in the s and p orbitals.

ddd [48]

Answer:

C 16

Explanation:

Since there are 6 valence electrons, this belongs in the same column as Oxygen. In fact, this is a diagram of a neutral oxygen atom with 8 total electrons.

7 0

3 years ago

Read 2 more answers

A sample of neon is collected at 2.7 atm and 12.0 oC. It has a volume of 2.25 L. What would be the volume of this gas at STP?

Keith_Richards [23]

Answer:

5.82 L

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,

V₁ = 2.25 L

V₂ = ?

P₁ = 2.7 atm

T₁ = 12 ºC

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (12 + 273.15) K = 285.15 K

At STP, Pressure = 1 atm and Temperature = 273.15 K

So,

P₂ = 1 atm

T₂ = 273.15 K

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac {{2.7}\times {2.25}}{285.15}=\frac {{1}\times {V_2}}{273.15}$

Solving for V₂ , we get:

<u>V₂ = 5.82 L</u>

7 0

4 years ago

A forest ecosystem starts with both Sonoran toads and Wood frogs. What would you expect to happen if the ecosystem becomes more

Gnom [1K]

Answer:

i think that sonoran toads would prosper and wood frogs would disappear

Explanation:

because of the provided information

5 0

3 years ago

In the laboratory a student combines 40.6 mL of a 0.113 M copper(II) sulfate solution with 26.4 mL of a 0.329 M copper(II) iodid

Arisa [49]

Answer : The final concentration of copper(II) ion is, 0.198 M

Explanation :

First we have to calculate the moles of $CuSO_4$ and $CuI_2$ .

$\text{Moles of }CuSO_4=\text{Concentration of }CuSO_4\times \text{Volume of solution}$

$\text{Moles of }CuSO_4=0.113mol/L\times 0.0406L=0.00459mol$

Moles of $CuSO_4$ = Moles of $Cu^{2+}$ = 0.00459 mol

and,

$\text{Moles of }CuI_2=\text{Concentration of }CuI_2\times \text{Volume of solution}$

$\text{Moles of }CuI_2=0.329mol/L\times 0.0264L=0.00869mol$

Moles of $CuI_2$ = Moles of $Cu^{2+}$ = 0.00869 mol

Now we have to calculate the total moles of copper(II) ion and total volume of solution.

Total moles copper(II) ion = 0.00459 mol + 0.00869 mol

Total moles copper(II) ion = 0.0133 mol

and,

Total volume of solution = 40.6 mL + 26.4 mL = 64.0 mL = 0.067 L (1 L = 1000 mL)

Now we have to calculate the final concentration of copper(II) ion.

$\text{Final concentration of copper(II) ion}=\frac{\text{Total moles}}{\text{Total volume}}$

$\text{Final concentration of copper(II) ion}=\frac{0.0133mol}{0.067L}=0.198mol/L=0.198M$

Thus, the final concentration of copper(II) ion is, 0.198 M

5 0

3 years ago