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2 years ago

What conditions on Earth allow life to thrive?

Chemistry

2 answers:

olga_2 [115]2 years ago

8 0

Answer:

atmosphere containing free oxygen --> atmosphere shielding against UV --> ozone layer in upper atmosphere --> axis of rotation generally perpendicular to orbital plane --> magnetic field that deflects the solar wind --> liquid water --> temperature near the triple point of water --> range of temperature sufficiently narrow, over appreciable portion of the planet's surface, conducive to the chemistry of life

satela [25.4K]2 years ago

7 0

Ugghhhhhhhhhhhhhhhhhhhhhhhhhh

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14.2 grams of Na2SO4 is dissolved in water to make a 2.50 L

Trava [24]

Answer:

0.04 M

Explanation:

Given data:

Mass of Na₂SO₄= 14.2 g

Volume of solution = 2.50 L

Molarity of solution = ?

Solution:

Number of moles of Na₂SO₄:

Number of moles = mass/ molar mass

Number of moles = 14.2 g/ 142.04 g/mol

Number of moles = 0.1 mol

Molarity :

Molarity = number of moles of solute / volume of solution in L

Molarity = 0.1 mol / 2.50 L

Molarity = 0.04 M

6 0

3 years ago

Which of the following below is the complete ionic equation of the reaction: Calcium nitrate and sodium sulfide solutions react

3241004551 [841]

To find the net ionic equation we must first write the balanced equation for the reaction. We must bear in mind that the reagents Ca(NO3)2 and Na2S are in the aqueous state and as product we will have CaS in the solid state, since it is not soluble in water and NaNO3 in the aqueous state.

The balanced equation of the reaction will be:

$Ca(NO_3)_{2(aq)}+Na_2S_{(aq)}\rightarrow CaS_{(s)}+2NaNO_{3(aq)}$

Ca(NO3)2(aq) + → Ca(aq) + 2Na(s)NO3Now, c(aq)ompounds in the aqueous state can be written in their ionic form, so the reaction will transform into:Na2S +

$Ca_{(aq)}^{2+}+2(NO_3)_{(aq)}^-+2Na_{(aq)}^++S_{(aq)}^{2-}\operatorname{\rightarrow}CaS_{(s)}+2Na_{(aq)}^++2NO^-_{3(aq)}$

So, the answer will be option A

5 0

1 year ago

What is bio diversity

denpristay [2]

Answer:

Biodiversity is the variety and variability of life on Earth. Biodiversity is typically a measure of variation at the genetic, species, and ecosystem level.

Explanation:

5 0

3 years ago

Read 2 more answers

A voltaic cell consists of a Zn>Zn2+ half-cell and a Ni>Ni2+ half-cell at 25 °C. The initial concentrations of Ni2+ and Zn

nlexa [21]

Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of $Ni^{2+}$ has fallen to 0.500 M is, 0.52 V

(c) The concentrations of $Ni^{2+}$ and $Zn^{2+}$ when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M

Explanation :

The values of standard reduction electrode potential of the cell are:

$E^0_{[Ni^{2+}/Ni]}=-0.23V$

$E^0_{[Zn^{2+}/Zn]}=-0.76V$

From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $Zn\rightarrow Zn^{2+}+2e^-$ $E^0_{[Zn^{2+}/Zn]}=-0.76V$

Reaction at cathode (reduction) : $Ni^{2+}+2e^-\rightarrow Ni$ $E^0_{[Ni^{2+}/Ni]}=-0.23V$

The balanced cell reaction will be,

$Zn(s)+Ni^{2+}(aq)\rightarrow Zn^{2+}(aq)+Ni(s)$

First we have to calculate the standard electrode potential of the cell.

$E^o=E^o_{cathode}-E^o_{anode}$

$E^o=E^o_{[Ni^{2+}/Ni]}-E^o_{[Zn^{2+}/Zn]}$

$E^o=(-0.23V)-(-0.76V)=0.53V$

(a) Now we have to calculate the cell potential.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(0.100)}{(1.50)}$

$E_{cell}=0.49V$

(b) Now we have to calculate the cell potential when the concentration of $Ni^{2+}$ has fallen to 0.500 M.

New concentration of $Ni^{2+}$ = 1.50 - x = 0.500

x = 1 M

New concentration of $Zn^{2+}$ = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

Now put all the given values in the above equation, we get:

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(1.1)}{(0.500)}$

$E_{cell}=0.52V$

(c) Now we have to calculate the concentrations of $Ni^{2+}$ and $Zn^{2+}$ when the cell potential falls to 0.45 V.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}+x]}{[Ni^{2+}-x]}$

Now put all the given values in the above equation, we get:

$0.45=0.53-\frac{0.0592}{2}\log \frac{(0.100+x)}{(1.50-x)}$

$x=1.49M$

The concentration of $Ni^{2+}$ = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of $Zn^{2+}$ = 0.100 + x = 0.100 + 1.49 = 1.59 M

5 0

3 years ago

5. Base your answer to the following question on the

elixir [45]

Answer:

B) 2Crº + 6e- --> 2Cr3+

Explanation:

The process of oxidation is where electrons are lost. Thus, out of the 2 ions that change charge(Cr and Cu), we must choose the one where the oxidation number increases(which means electrons are lost). Cr goes from an oxidation number of 0 to an oxidation number of 3+, while Cu goes from an oxidation number of 2+ to 0. Thus, we are looking at the half reaction for Cr. Half reactions never have subtracting electrons, so the answer must be B. I am assuming that last plus should be a -->

4 0

3 years ago

Read 2 more answers