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3 years ago

8

A spring-fed stream forms a pond that contains watercress plants, minnows, and crayfish. The pond receives 5 hours of direct sun

light every day. Which of the following is an abiotic factor in this ecosystem?
the pond water

the watercress plants

the minnows

the crayfish

2 answers:

Snowcat [4.5K]3 years ago

8 0

Answer:

The pond water

d1i1m1o1n [39]3 years ago

3 0

The pond water because of the water cycles

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Does a displacement reaction take place in 'magnesium + lead nitrate'? and if so why?

cestrela7 [59]

Answer:

Maybe or maybe not (not sure)

Explanation:

A displacement reaction is a type of reaction where one element is displaced by another from a compound.

In the case of magnesium and lead nitrate, magnesium is more reactive than lead. Therefore, it will displace lead from lead nitrate to form magnesium nitrate and lead.

The reaction can be represented as:

Mg(s) + Pb(NO3)2(aq) → Mg(NO3)2(aq) + Pb(s)

Another answer could be;

A displacement reaction does not take place in 'magnesium + lead nitrate' because magnesium is more reactive than lead.

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They flow through the inslation of the house .

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3 years ago

Which of these is a benefit of fuel cell cars?

Otrada [13]

Answer:

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Explanation:

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What is the percent error when a student measures the volume to be 19.3 liters when

katrin2010 [14]

Answer:

The answer is

<h2>13.84 %</h2>

Explanation:

The percentage error of a certain measurement can be found by using the formula

$P(\%) = \frac{error}{actual \: \: number} \times 100\% \\$

From the question

actual volume = 22.4 L

error = 22.4 - 19.3 = 3.1

The percentage error is

$P(\%) = \frac{3.1}{22.4} \times 100 \\ = 13.839285714...$

We have the final answer as

<h3>13.84 %</h3>

Hope this helps you

7 0

3 years ago

Consider the reaction. 2 HBr(g) ¡ H2(g) + Br2(g) a. Express the rate of the reaction in terms of the change in concentration of

Studentka2010 [4]

Answer :

(A) The rate expression will be:

$Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}$

(B) The average rate of the reaction during this time interval is, 0.00176 M/s

(C) The amount of Br₂ (in moles) formed is, 0.0396 mol

Explanation :

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The given rate of reaction is,

$2HBr(g)\rightarrow H_2(g)+Br_2(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }HBr=-\frac{1}{2}\frac{d[HBr]}{dt}$

$\text{Rate of disappearance of }H_2=+\frac{d[H_2]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{d[Br_2]}{dt}$

<u>Part A:</u>

The rate expression will be:

$Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}$

<u>Part B:</u>

$\text{Average rate}=-\frac{1}{2}\frac{d[HBr]}{dt}$

$\text{Average rate}=-\frac{1}{2}\frac{(0.512-0.600)M}{(25.0-0.0)s}$

$\text{Average rate}=0.00176M/s$

The average rate of the reaction during this time interval is, 0.00176 M/s

<u>Part C:</u>

As we are given that the volume of the reaction vessel is 1.50 L.

$\frac{d[Br_2]}{dt}=0.00176M/s$

$\frac{d[Br_2]}{15.0s}=0.00176M/s$

$[Br_2]=0.00176M/s\times 15.0s$

$[Br_2]=0.0264M$

Now we have to determine the amount of Br₂ (in moles).

$\text{Moles of }Br_2=\text{Concentration of }Br_2\times \text{Volume of solution}$

$\text{Moles of }Br_2=0.0264M\times 1.50L$

$\text{Moles of }Br_2=0.0396mol$

The amount of Br₂ (in moles) formed is, 0.0396 mol

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3 years ago