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2 years ago

Which type of ecological succession has occurred? Could it be both? Why?

1 answer:

4 0

Two different types of succession—primary and secondary—have been distinguished. Primary succession occurs in essentially lifeless areas—regions in which the soil is incapable of sustaining life as a result of such factors as lava flows, newly formed sand dunes, or rocks left from a retreating glacier,I hope that’s correct

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Some fruits and vegetables are preserved by pickling them. Nandini got confused

stepan [7]

Answer:

Explanation:

3 0

2 years ago

Graphs are a way to present data in picture form t or f

Irina-Kira [14]

I’m pretty sure it’s true

6 0

3 years ago

20 mL of Ba(OH)2 solution with unknown concentration was neutralized by the addition of 43.89 mL of a .1355 M HCl solution. Calc

bezimeni [28]

Answer:

Concentration of the barium ions = $[Ba^{2+}] = 0.4654 M$

Concentration of the chloride ions = $[Cl^{-}]=0.9308 M$

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of hydrogen chloride = n

Volume of hydrogen chloride solution = 43.89 mL = 0.04389 L

Molarity of the hydrogen chloride = 0.1355 M

$n=0.1355 M\times 0.04389 L=0.005947 mol$

$Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O$

According to reaction, 2 moles of HCl reacts with 1 mole of barium hydroxide.

Then 0.05947 moles of HCl will react with:

$\frac{1}{2}\times 0.05947 mol=0.029735 mol$ barium hydroxide

Moles of barium hydroxide = 0.029735 mol

$Ba(OH)_2(aq)\rightarrow Ba^{2+}(aq)+2OH^-(aq)$

1 mole of barium hydroxide gives 1 mole of barium ion in an aqueous solution. Then 0.029735 moles of barium hydroxide will give:

$=1\times 0.029735 mol= 0.029735 mol$

Volume of solution after neutralization reaction :

= 20.0 mL + 43.89 mL = 63.89 mL = 0.06389 L

Concentration of the barium ions = $[Ba^{2+}]$

$[Ba^{2+}]=\frac{0.029735 mol}{0.06389 L}=0.4654 M$

$Ba(Cl)_2(aq)\rightarrow Ba^{2+}(aq)+2Cl^-(aq)$

1 mole of barium chloride gives 1 mole of barium ions and 2 moles of chloride ions in an aqueous solution.

Then concentration of chloride ions will be:

$[Cl^-]=2\times [Ba^{2+}]=2\times 0.4654 M=0.9308 M$

8 0

3 years ago

If an element has 2+ valence electrons, does it transfer only one or more than one valence electrons

just olya [345]

Answer:

element having 2+ valence electrons can transfer its more than one electron that is 2 electron completely.

Explanation:

Group IIA have 2+ valency and two electrons in its valance shell.

Its Electropositivity is high and have the tendency to donate it two electrons.

Element of IIA form ionic with most electronegative element.

Examples:

Cu²⁺, Mg²⁺, Sr²⁺ are examples having 2+ valance electron

one of the following is examples of element that have 2+ valence electrons

MgCl₂

Atomic number of Magnesium (Mg) is 12

Electronic Configuration of Mg:

1s², 2s², 2p⁶, 3s²

K =2

L = 8

M = 2

So, it have to give its 2 electrons to form a stable compound.

Similarly

Chlorine atomic number is 17

Electronic Configuration of Chlorine:

1s², 2s², 2p⁶, 3s², 3p⁵

K =2

L = 8

M = 7

So, it have to gain one electrons to form a stable compound and complete its octet.

So,

Two chlorine atom as a molecule gain 2 electrons from Mg²⁺ atom

So one Mg²⁺ and 2 Cl⁻ atoms form an ionic bond

where in this ionic bond Mg²⁺ transfer its 2 valence electron completely and chlorine molecule accept 2 electrons.

Cl-----Mg------Cl

So the Answer is

element having 2+ valence electrons can transfer its more than one electron that is 2 electron completely.

8 0

2 years ago

The half life for the decay of carbon-14 is 5.73 x 10 years. Suppose the activity due to the radioactive decay of the carbon-14

Elena-2011 [213]

Answer:

Age of the atifact is $4.2\times 10^{2}$ years

Explanation:

For first -order radioactive decay- $A_{t}=A_{0}(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$
$A_{t}$ represents activity of radioactive nuclide after t time, $A_{0}$ represents initial activity of radioactive nuclide and $t_{\frac{1}{2}}$ represents half-life
Here, $A_{t}=19Bq$ , $A_{0}=20Bq$ and $t_{\frac{1}{2}}=5.73\times 10^{3}years$