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2 years ago

Write a hypothesis about the effect of dry conditions on earthworm behavior. Use the "if . . . then . . .

Chemistry

2 answers:

alexgriva [62]2 years ago

5 0

Answer:

If the earthworm's skin dries then they will not breath properly because moisture help them out to ease the diffusion of oxygen through its skin.

Explanation:

Hello,

Based on the required hypothesis, the best describing one considering the question "How is earthworm behavior affected by external stimuli?", would be:

If the earthworm's skin dries then they will not breath properly because moisture help them out to ease the diffusion of oxygen through its skin.

In this case, we are specifying a possible condition, this is "the earthworm's skin dries" which is a cause of an external stimuli (dryness). Next, we propose an effect, this is "they will not breath properly", that is caused by the initial part. Finally, we explain the effect via the cause "moisture help them out to ease the diffusion of oxygen through its skin" because is something testable via the subsequent experiments when considering the scientific method.

Best regards.

Vlada [557]2 years ago

4 0

sorry about the late response...

<u>If an earthworm is exposed to dry conditions, then it will retreat to a moist place because its skin needs to stay moist for the earthworm to survive.</u>

Guest

2 years ago

who u

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A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 417 pm. The diameter of the metal atom is:

enyata [817]

Answer:

b. 295 pm

Explanation:

To answer this question we need to use the equation of a face-centered cubic laticce:

Edge length = √8 R

<em>Where R is radius of the atom.</em>

<em />

Replacing:

417pm = √8 R

R = 147.4pm is the radius of the atom

As diameter = 2 radius.

Diameter of the metal atom is:

147.4pm* 2 =

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2 years ago

A transition in the balmer series for hydrogen has an observed wavelength of 434 nm. Use the Rydberg equation below to find the

DanielleElmas [232]

Answer:

i. n = 5

ii. ΔE = 7.61 × $10^{-46}$ KJ/mole

Explanation:

1. ΔE = (1/λ) = -2.178 × $10^{-18}$ ( $\frac{1}{n^{2}_{final} }$ - $\frac{1}{n^{2}_{initial} }$ )

(1/434 × $10^{-9}$ ) = -2.178 × $10^{-18}$ ( $\frac{n^{2}_{initial} - n^{2}_{final} }{n^{2}_{final} n^{2}_{initial} }$ )

⇒ 434 × $10^{-9}$ = (1/-2.178 × $10^{-18}$ ) $\frac{n^{2}_{final} *n^{2}_{initial} }{n^{2}_{initial} - n^{2}_{final} }$

But, $n_{final}$ = 2

434 × $10^{-9}$ = (1/2.178 × $10^{-18}$ ) $\frac{2^{2} n^{2}_{initial} }{n^{2}_{initial} - 2^{2} }$

434 × $10^{-9}$ × 2.178 × $10^{-18}$ = $(\frac{4n^{2}_{initial} }{n^{2}_{initial} - 4 })$

⇒ $n_{initial}$ = 5

Therefore, the initial energy level where transition occurred is from 5.

2. ΔE = hf

= (hc) ÷ λ

= (6.626 × 10−34 × 3.0 × $10^{8}$ ) ÷ (434 × $10^{-9}$ )

= (1.9878 × $10^{-25}$ ) ÷ (434 × $10^{-9}$ )

= 4.58 × $10^{-19}$ J

= 4.58 × $10^{-22}$ KJ

But 1 mole = 6.02× $10^{23}$ , then;

energy in KJ/mole = (4.58 × $10^{-22}$ KJ) ÷ (6.02× $10^{23}$ )

= 7.61 × $10^{-46}$ KJ/mole

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2 years ago

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The Charge Of Cesium Is +1

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2. If 2.50g of sodium hydroxide is being reacted with 4.30g of magnesium chloride, how many grams of magnesium hydroxide would b

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Answer:

1.822 g of magnesium hydroxide would be produced.

Explanation:

Balanced reaction: $2NaOH+MgCl_{2}\rightarrow Mg(OH)_{2}+2NaCl$

Compound Molar mass (g/mol)

NaOH 39.997

$MgCl_{2}$ 95.211

$Mg(OH)_{2}$ 58.3197

So, 2.50 g of NaOH = $\frac{2.50}{39.997}$ mol of NaOH = 0.0625 mol of NaOH

4.30 g of $MgCl_{2}$ = $\frac{4.30}{95.211}$ mol of $MgCl_{2}$ = 0.0452 mol of $MgCl_{2}$

According to balanced equation-

2 mol of NaOH produce 1 mol of $Mg(OH)_{2}$

So, 0.0625 mol of NaOH produce $(\frac{0.0625}{2})$ mol of NaOH or 0.03125 mol of NaOH

1 mol of $MgCl_{2}$ produces 1 mol of $Mg(OH)_{2}$

So, 0.0452 mol of $MgCl_{2}$ produce 0.0452 mol of $Mg(OH)_{2}$

As least number of moles of $Mg(OH)_{2}$ are produced from NaOH therefore NaOH is the limiting reagent.

So, amount of $Mg(OH)_{2}$ would be produced = 0.03125 mol

= $(0.03125\times 58.3197)$ g

= 1.822 g

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