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3 years ago

9

A fox hunts a rabbit. When the fox captures the rabbit, he consumes it. What is the primary reason for this organism interaction

?
A. nourishment

B. companionship

C. reproduction

D. protection

2 answers:

Umnica [9.8K]3 years ago

4 0

it is nourishment trust me.

jolli1 [7]3 years ago

3 0

I'm pretty sure it's not B or D.

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Calculate the moles of calcium oxalate formed from 5 grams of potassium oxalate (molar mass 184.24 g/mol) the mole mole ratio fr

Nataly_w [17]

5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.

Calcium oxalate (CaC₂O₄) is obtained by the reaction of 5 g of potassium oxalate (K₂C₂O₄).

We can calculate the moles of CaC₂O₄ obtained considering the following relationships.

The molar mass of K₂C₂O₄ is 184.24 g/mol.
The mole ratio of K₂C₂O₄ to CaC₂O₄ is 2:1.

$5 g K_2C_2O_4 \times \frac{1molK_2C_2O_4}{184.24gK_2C_2O_4} \times \frac{1molCaC_2O_4}{2molK_2C_2O_4} = 0.03molCaC_2O_4$

5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.

Learn more: brainly.com/question/15288923

7 0

2 years ago

How many moles of oxygen are needed to burn 425g of sulfur

Anika [276]

I believe 212.5m, but I may be wrong, I’m a little rusty with moles

3 0

3 years ago

Read 2 more answers

1. Calculate the mass (g), of 4.5 atoms of oxygen.<br> I

konstantin123 [22]

Answer

thus, 4 moles of oxygen gas (O2) would have a mass of 128 g.

3 0

3 years ago

Help in my exam plzz now

swat32

Answer:

yes your answer is correct

I hope it helps mate

I will always help you understanding your assingments have a great day

#Captainpower:)

8 0

3 years ago

How many grams of sulfuric acid is needed to neutralize 380 ml of solution with pH = 8.94

erma4kov [3.2K]

Answer : The mass of sulfuric acid needed is $16.23\times 10^{-5}g$ .

Solution : Given,

pH = 8.94

Volume of solution = 380 ml = $380\times 10^{-3}$ $(1ml=10^{-3}L)$

Molar mass of sulfuric acid = 98.079 g/mole

As we know,

$pH+pOH=14\\pOH=14-8.94=5.06$

$pOH=-log[OH^-]$

$5.06=-log[OH^-]$

$[OH^-]=0.00000871=8.71\times 10^{-6}mole/L$

Now we have to calculate the moles of $OH^-$ .

Formula used : $Moles=Concentration\times Volume$

$\text{ Moles of }[OH^-]=\text{ Concentration of }[OH^-]\times Volume\\\text{ Moles of }[OH^-]=(8.71\times 10^{-6}mole/L)\times (380\times 10^{-3}L)=3309.8\times 10^{-9}moles$

For neutralization, equal number of moles of $H^+$ ions will neutralize same number of $OH^-$ ions.

$\text{ Moles of }[OH^-]=\text{ Moles of }[H^+]=3309.8\times 10^{-9}moles$

As, $H_2SO_4\rightarrow 2H^++SO^{2-}_4$

From this reaction, we conclude that

2 moles of $H^+$ ion is given by the 1 mole of $H_2SO_4$

$3309.8\times 10^{-9}$ moles of $H^+$ ion is given by $\frac{3309.8\times 10^{-9}}{2}=1654.9\times 10^{-9}$ moles of $H_2SO_4$

Now we have to calculate the mass of sulfuric acid.

Mass of sulfuric acid = Moles of $H_2SO_4$ × Molar mass of sulfuric acid

Mass of sulfuric acid = $(1654.9\times 10^{-9}moles)\times (98.079g/mole)=162310.94\times 10^{-9}=16.23\times 10^{-5}g$

Therefore, the mass of sulfuric acid needed is $16.23\times 10^{-5}g$ .

3 0

3 years ago