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2 years ago

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An experiment was conducted to measure the effect of equal amounts of fertilizer on the growth of bean plants and corn plants. T

he results indicated that, although both seedlings benefited from the fertilizer, the bean plants benefited slightly more. Which graph best shows these results?
A. X
B. W
C. Z
D. Y

2 answers:

podryga [215]2 years ago

8 0

Answer:

A

Explanation:

did it on study island

ss7ja [257]2 years ago

3 0

Answer:

b

Explanation:

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Put the list in chronological order (1–5).

Leokris [45]

Explanation:

Filtration is a separation technique in which solid particles suspended in liquid medium are separated by allowing the mixture through the pores of the filter paper. By this solid particles get collect on filter paper and liquid drains out from the pores of the filter paper.

The chronological order for given steps will be:

Weigh and fold the filter paper.
Place the filter paper in the funnel, then place the funnel in the Erlenmeyer flask.
Allow the solid/liquid mixture to drain through the filter.
Use water to rinse the filter paper containing the mixture.
Weigh the dried filter paper and copper.

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2 years ago

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Much to everyone’s surprise, nitrogen monoxide (NO) has been found to act as a neurotransmitter. To prepare to study its effect,

SVEN [57.7K]

Answer: 0.0043mole

Explanation:Please see attachment for explanation

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2 years ago

Which reactants would lead to a spontaneous reaction?

balandron [24]

Answer: Option (b) is the correct answer.

Explanation:

The elements which have excess or deficiency of electrons will react readily.

Atomic number of Mn is 25 and electronic configuration of $Mn^{2+}$ is [Ar] $4s^{0}3d^{5}$ . This configuration is stable.

Atomic number of Cr is 24 and electronic configuration of $Cr$ is [Ar] $4s^{1}3d^{5}$ . This configuration is not stable.

Atomic number of Fe is 26 and electronic configuration of $Fe$ is [Ar] $4s^{2}3d^{6}$ . This configuration is stable.

Atomic number of Cu is 29 and electronic configuration of $Cu^{2+}$ is [Ar] $4s^{0}3d^{9}$ . This configuration is not stable.

Atomic number of Al is 13 and electronic configuration of Al is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}$ . This configuration is not stable.

Atomic number of Ba is 56 and electronic configuration of $Ba^{2+}$ is [Kr] $4d^{10}5s^{2}5p^{6}$ . This configuration is stable.

Atomic number of Mg is 12 and electronic configuration of $Mg^{2+}$ is $1s^{2}2s^{2}2p^{6}$ . This configuration is stable.

Atomic number of Sn is 50 and electronic configuration of Sn is [Kr] $4d^{10}5s^{2}5p^{2}$ . This configuration is stable.

Thus, we can conclude that out of the given options, only Fe and $Cu^{2+}$ reactants would lead to a spontaneous reaction as they have incomplete sub-shells. Therefore, in order to gain stability they will readily react.

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2 years ago

Calculate the molar mass of a gas that diffuses three times faster than oxygen under similar conditions.

JulsSmile [24]

<span> rate 3/1= square root of 32/x
square both sides
9/1=32x
x = 32/9
= 3.6
Must be He
molar mass =4
</span>

7 0

2 years ago

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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⁺.

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2 years ago