All categories

3 years ago

Which character of Life distinguishes a fire and a flowering plant

Chemistry

1 answer:

elena55 [62]3 years ago

7 0

Answer:

a flowering plant needs water while a fire doesn't.

That what I guessed???

You might be interested in

How many molecules of ethane are present in 64.28 liters of ethane gas (C2H6) at STP

Rom4ik [11]

Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is, $17.277\times 10^{23}$ molecule.

Solution :

At STP,

22.4 L volume of ethane present in 1 mole of ethane gas

64.28 L volume of ethane present in $\frac{64.28L}{22.4L}\times 1mole=2.869moles$ of ethane gas

And, as we know that

1 mole of ethane molecule contains $6.022\times 10^{23}$ molecules of ethane

2.869 moles of ethane molecule contains $2.869\times 6.022\times 10^{23}=17.277\times 10^{23}$ molecules of ethane

Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is, $17.277\times 10^{23}$ molecule.

4 0

3 years ago

Read 2 more answers

Which of these groups CANNOT harvest energy from the sun?

Tomtit [17]

Viruses..................

8 0

3 years ago

Read 2 more answers

Fr33 points ;p have a good day everyone

Doss [256]

Answer:

thx

Explanation:

have a blessed day can i have brainlyest?

5 0

3 years ago

Scientist who discovered the atomic nucleus is ________. dalton b. thomson chadwick d. rutherford

german

It’s D. Rutherford discovered the atomic nucleus.

7 0

3 years ago

Sodium phosphate is added to a solution that contains 0.0030 M aluminum nitrate and 0.016 M calcium chloride. The concentration

gavmur [86]

Explanation:

It is given that aluminium nitrate and calcium chloride are mixed together with sodium phosphate.

And, $K_{sp} of AlPO_{4} = 9.84 \times 10^{-21}$

$K_{sp} of Ca_{3}(PO_{4})_{2} = 2.0 \times 10^{-29}$

Let us assume that the solubility be "s". And, the reaction equation is as follows.

$AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}$

$9.84 \times 10^{-21} = s \times s$

s = $9.92 \times 10^{-11}$

Also, $Ca_{3}(PO_{4})_{2} \rightleftharpoons 3Ca^{2+} + 2PO^{3-}_{4}$

$2 \times 10^{-29} = (3s)^{3} \times (2s)^{2}$

s = $7.14 \times 10^{-7}$

This means that first, aluminium phosphate will precipitate.

Now, we will calculate the concentration of phosphate when calcium phosphate starts to precipitate out using the $K_{sp}$ expression as follows.

$K_{sp} = [Ca^{2+}]^{3}[PO^{3-}_{4}]^{2}$

$2.0 \times 10^{-29} = (0.016)^{3}[PO^{3-}_{4}]^{2}$

$2.0 \times 10^{-29} = 4.096 \times 10^{-6} \times [PO^{3-}_{4}]^{2}$

$[PO^{3-}_{4}]^{2}$ = $4.88 \times 10^{-24}$

= $2.21 \times 10^{-12}$ M

Similarly, calculate the concentration of aluminium at this concentration of phosphate as follows.

$AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}$

$K_{sp} = [Al^{3+}][PO^{3-}_{4}]$

$9.84 \times 10^{-21} = [Al^{3+}] \times 2.21 \times 10^{-12}$

$[Al^{3+}] = 4.45 \times 10^{-9}$ M

Thus, we can conclude that concentration of aluminium will be $4.45 \times 10^{-9}$ M when calcium begins to precipitate.

7 0

3 years ago

Which character of Life distinguishes a fire and a flowering plant​

Which character of Life distinguishes a fire and a flowering plant