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2 years ago

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The data table shows the number of pumpkin seeds that germinate at different temperatures. What is the median number of seeds th

at germinate at 24C?
A. 80.5 B. 82 C. 78.5 D. 80

1 answer:

allochka39001 [22]2 years ago

7 0

Answer:

Explanation:

80

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The reaction below has an equilibrium constant kp=2.2×106 at 298 k. 2cof2(g)⇌co2(g)+cf4(g) calculate kp for the reaction below.

AnnZ [28]

<span>I believe the correct 2nd reaction is:</span>

cof2(g)⇌1/2 co2(g)+1/2 cf4(g)

where we can see that it is exactly one-half of the original

Therefore the new Kp is:

new Kp = (old Kp)^(1/2)

new Kp = (2.2 x 10^6)^(1/2)

<span>new Kp = 1,483.24 </span>

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You see a structural formula in which the symbols for elements are connected by a long dash. You can assume that the chemical bo

Shkiper50 [21]

<span>You see a structural formula in which the symbols for elements are connected by a long dash. I can assume that the chemical bonds in the compound are covalent.</span>

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2 years ago

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Some versions of the periodic table show hydrogen at the top of Group 1A(1) and at the top of Group 7A(17). What properties of h

lawyer [7]

Answer:Hydrogen is placed such because it exhibits some similar characteristics of both group1 and group VII elements.

Explanation:

The reason why hydrogen is similar to group 1 metals:

#It has same valence electron and inorder achieve octet state it can lose that electron and forms H+ ion

#It acts as a good reducing agent similar to group1 metals

#It can also halides

Similarity to halogens:

#hydrogen can also gain one electron to gain noble gas configuration. It can combine with other non metals to form molecules with covalent bonding.

#It exists as diatomin molecule,H2

#Have the same electronegativity nature

#its reaction with other metal

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2 years ago

Carbon monoxide and molecular oxygen react to form carbon dioxide. A 50.0 L reactor at 25.0 oC is charged with 1.00 bar of CO. T

Umnica [9.8K]

Answer:

CO = zero

CO2 =1 bar

O2 = 2.02 bar

Explanation:

We are given

initial pressure of CO = 1bar

total pressure = 3.52 bar

so initial pressure of O2 = 3.52 - 1 = 2.52 bar

the reaction is

2CO + O2 → 2CO2

using the unitary method

2 moles of CO2 → 1 mole of O2

1 bar of CO → $\frac{1}{2} * 1= 0.5 bar$ (required)

but we have more oxygen present , that means CO is the limiting reagent

Final pressure of CO will be zero as it is the limiting reagent so it will be consumed completely
1 bar of CO → $\frac{2mol CO2}{2mol CO} * 1= 1 bar$ of CO2
2.52 bar O2 (initially) - 0.5 bar (reacted) = 2.02bar O2

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2 years ago

If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th

fredd [130]

Answer:

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

A)

We assume that s to be the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

[Pb²⁺] = s

Then [I⁻] = 2s

$K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} = 4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

B)

The Concentration of Pb²⁺ in water is calculated as :

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

$\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}$

$\mathbf{s} =\sqrt[3]{1.75*10^{-9}}$

$\mathbf{s} =\mathbf{1.21*10^{-3} \ mol/L }$

The Concentration of Pb²⁺ in 1.0 mol·L⁻¹ NaI

$\mathbf{PbCl{_2}} \leftrightharpoons \ \ \ \ \ \ \ \mathbf{Pb^{2+}} \ \ \ \ \ + \ \ \ \ \ \ \ \mathbf{2 I^-}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+2x}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0+2x}$

The equilibrium constant:

$K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \ m/L$

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

3 0

2 years ago