All categories

2 years ago

Many food chains together create a______

Chemistry

1 answer:

Amanda [17]2 years ago

3 0

Many food chains together create a food web.

hope this helps,, have a nice day :))

You might be interested in

Which of the following is true of the water molecule?

Yuki888 [10]

The Answer To This Is B.

7 0

3 years ago

Which value of x from the set {4, 5, 6, 7), makes this equation true?

irina [24]

C and d is the answer

3 0

3 years ago

Read 2 more answers

How do i find the mole of something?

nikklg [1K]

To find the mole of a substance, you take the mass divided by the molar mass of that substance~

3 0

3 years ago

Read 2 more answers

ch question carries 2 mark. Time Remaining : 00 : 46 : 33 Some oxides are given below. (i)Na2O (ii)NO2 (iii) CO2 (iv) MgO a) Whi

cluponka [151]

<h3>Further explanation</h3>

The basic oxide is an oxide-forming a base solution.

These oxides are mainly from group 1 alkaline and group 2-alkaline earth

If this oxide is dissolved in water it will form an alkaline solution

LO + H₂O --> L(OH)₂ ---> alkaline earth

L₂O + H₂O --> LOH --> alkaline

So the basic oxides : Na₂O and MgO

Na₂O + H₂O --> NaOH (sodium hydroxide, strong base)

MgO + H₂O --> Mg(OH)₂ (magnesium hydroxide, strong base)

The aqueous solution of CO₂ , obtained by dissolving CO₂ in water

CO₂ + H₂O --> H₂CO₃ (carbonic acid)

In general, basic oxide is obtained from metal oxide, while acid oxide is obtained from non-metal oxide

6 0

3 years ago

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago