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3 years ago

10

Which ocean zone lies inside V-shaped ocean trenches? O A. Bathypelagic O B. Abyssopelagic Ο Ο Ο Ο O C. Mesopelagic O D. Hadalpe

lagic

1 answer:

alisha [4.7K]3 years ago

5 0

Answer:

O D. Hadalpelagic is the answer.

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What is the volume of 68.0 g of ether if the density of ether is 0.72 g/mL?

umka2103 [35]

Answer:

94.44

Explanation:

Volume is equal to Mass/Density so therefore, you do the mass which is 68.0 g/0.72 g/mL which is the density and get 94.44 mL because the g cancel each other out when it comes to the label!

3 0

3 years ago

CaC12 + NaCO3–>CaCO3 +NaC1

lapo4ka [179]

I think it would be c

7 0

3 years ago

Fe3+(aq) (yellow) + SCN-(aq) (colorless) FeSCN2+(aq) (blood-red) Chloride ions are colorless. Potassium ions are also colorless.

harina [27]

Explanation:

$Fe^{3+}\text{(aq)(yellow)}+SCN^-\text{(aq)(colorless)}\rightleftharpoons FeSCN^{2+}\text{(aq)(blood-red)}$

$K^+\text{(aq)(colorless)}+Cl^-\text{(aq)(colorless)}\rightarrow KCl$

The above two reactions xcan also be written in form of single chemcial equation:

$FeCl_3+K(SCN)\rightleftharpoons KCl+[Fe(SCN)]Cl_2(blood-red)$

1. Color of ferric chloride solution is yellow. This is due to presence of ferric ions which have yellow color in their aqueous solutions.

2. KSCN has the colorless solution. This due to potassium ion forms colorless aqueous solution.

3. On mixing, KSCN with $FeCl_3$ we will get blood red color solution of $[FeSCN]Cl_2$ .

5 0

3 years ago

The equilibrium constant, Kc, for the following reaction is 1.29E-2 at 600 K. COCl2(g) CO(g) Cl2(g) Calculate the equilibrium co

RSB [31]

Answer:

At Equilibrium

[COCl₂] = 0.226 M

[CO] = 0.054 M

[Cl₂] = 0.054 M

Explanation:

Given that;

equilibrium constant Kc = 1.29 × 10⁻² at 600k

the equilibrium concentrations of reactant and products = ?

when 0.280 moles of COCl2(g) are introduced into a 1.00 L vessel at 600 K. [COCl²]

Concentration of COCl₂ = 0.280 / 1.00 = 0.280 M

COCl₂(g) ----------> CO(g) + Cl₂(g)

0.280 0 0 ------------ Initial

-x x x

(0.280 - x) x x ----------- equilibrium

we know that; solid does not take part in equilibrium constant expression

so

KC = [CO][Cl₂] / COCl₂

we substitute

1.29 × 10⁻² = x² / (0.280 - x)

0.0129 (0.280 - x) = x²

x² = 0.003612 - 0.0129x

x² + 0.0129x - 0.003612 = 0

x = -b±√(b² - 4ac) / 2a

we substitute

x = [-(0.0129)±√((0.0129)² - 4×1×(-0.003612))] / [2 × 1 ]

x = [-0.0129 ± √( 0.00017 + 0.01445)] / 2

x = [-0.0129 ± 0.1209] / 2

Acceptable value of x =[ -0.0129 + 0.1209] / 2

x = 0.108 / 2

x = 0.054

At equilibrium

[COCl₂] = (0.280 - x) = 0.280 - 0.054 = 0.226 M

[CO] = 0.054 M

[Cl₂] = 0.054 M

7 0

3 years ago

Use molecular orbital theory to determine whether He2 2+ or He2 + is more stable. Use molecular orbital theory to determine whet

julia-pushkina [17]

Answer:

The He₂ 2+ ion is more stable since it has a higher bond order (bond order = 1) than the He₂ + ion (bond order = 1/2).

Explanation:

Molecular orbital of He₂⁺

$1\sigma_{1s}^21\sigma(star)_{1s}^1$

There are two electrons in bonding and 1 electron in antibonding orbital

Bond order = $\frac{(2-1)}{2}$

= $\frac{1}{2}$

Molecular orbital of He₂⁺²

$1\sigma_{1s}^21\sigma(star)_{1s}^0$

There are two electrons in bonding and 0 electron in antibonding orbital

Bond order = $\frac{(2-0)}{2}$

= 1

So bond order of He₂⁺² is 1 which is more stable than He₂⁺ whose bond order is $\frac{1}{2}$ .

7 0

3 years ago