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3 years ago

13

Tertiary or top consumers are animals at the top of the food web. They may feed off either primary or secondary consumers. What

animals would fit into this classification?

1 answer:

babymother [125]3 years ago

7 0

Answer:

Coyotes, Bald Eagles, & Gray wolf.

You might be interested in

Whats the number of moles of O2(g) needed to completely react with 8 moles of CO(g).

slava [35]

A mole of CO2 = 2 moles of O2
8 CO moles x 2 =

16 moles

7 0

3 years ago

True or False: The Ring of Fire is a section along the Atlantic Ocean where you have active volcanoes

Romashka [77]

False because the Atlantic Ocean dose not have active volcanoes besides that’s a ocean so no

6 0

2 years ago

Read 2 more answers

3. How many moles of silver is 8.46x1024 atoms of<br> silver?

ioda

Answer:

<h2>14.05 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{8.46 \times {10}^{24} }{6.02 \times {10}^{23} } \\ = 14.053156$

We have the final answer as

<h3>14.05 moles</h3>

Hope this helps you

6 0

2 years ago

An iron ore sample weighing 0.5562 g is dissolved HCl (aq), and the iron is obtained as Fe2 in solution. This solution is then t

erik [133]

Answer:

$\% Fe^{+2}=70%$

Explanation:

Hello,

In this case, we could considering this as a redox titration:

$Fe^{+2}+(Cr_2O_7)^{-2}\rightarrow Fe^{+3}+Cr^{+3}$

Thus, the balance turns out (by adding both hydrogen ions and water):

$Fe^{+2}\rightarrow Fe^{+3}+1e^-\\(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow 2Cr^{+3}+7H_2O\\\\6Fe^{+2}+(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow Cr^{+3}+7H_2O+6Fe^{+3}+6e^-$

Thus, by stoichiometry, the grams of Fe+2 ions result:

$m_{Fe^{+2}}=0.04021\frac{molK_2Cr_2O_7}{L}*0.02872L*\frac{6molFe^{+2}}{1molK_2Cr_2O_7}*\frac{56gFe^{+2}}{1molFe^{+2}}=0.388gFe^{+2}$

Finally, the mass percent is:

$\% Fe^{+2}=\frac{0.388g}{0.5562g}*100\%\\ \% Fe^{+2}=70%$

Best regards.

8 0

2 years ago

Which is the limiting reactant if we start with 30.0 g Al and 30.0 h O2

lorasvet [3.4K]

Al

Explanation:

The limiting reactant will be Al:

4Al + 3O₂ → 2Al₂O₃

The limiting reactant is the reactant in short supply in a chemical reaction.

Given parameters:

Mass of Al = 30g Molar mass = 27g/mol

Number of moles = $\frac{mass}{molar mass}$ = $\frac{30}{27}$

Number of moles of Al = 1.111 mole

Mass of O₂ = 30g, molar mass = 32g/mol

Number of moles = $\frac{30}{32}$ = 0.94mol

In the reaction:

4 moles of Al reacted with 3 moles of O₂

1.11moles of Al will require $\frac{1.11 x3}{4}$ = 0.83mole to react

But we have been given 0.94mole of O₂. This is more than required.

Therefore O₂ is in excess and Al is the limiting reactant.

Learn more:

Limiting reagents brainly.com/question/6078553

#learnwithBrainly

4 0

2 years ago