Chang described two substances which participate in photosynthesis.

What charge does an ion have if it has more protons than electrons?

sergiy2304 [10]

I believe it has a positive charge, an easy way I remember is Positive and Proton both start with “P”

7 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

How much heat energy would be needed to raise the temperature of a 223 g sample of aluminum [(C=0.895 Jig Cy from 22.5°C to 55 0

dsp73

Answer : The heat energy needed would be, 6486.5125 J

Explanation :

To calculate the change in temperature, we use the equation:

$q=mc\Delta T\\\\q=mc(T_2-T_1)$

where,

q = heat needed = ?

m = mass of aluminum = 223 g

c = specific heat capacity of aluminum = $0.895J/g^oC$

$\Delta T$ = change in temperature

$T_1$ = initial temperature = $22.5^oC$

$T_2$ = final temperature = $55.0^oC$

Putting values in above equation, we get:

$q=223g\times 0.895J/g^oC\times (55.0-22.5)^oC$

$q=6486.5125J$

Therefore, the heat energy needed would be, 6486.5125 J

5 0

3 years ago

ANSWER ASAP GIVING BRAINIEST FIVE STARS AND HEART!

tino4ka555 [31]

I can describe one prop. of energy: it can not be created or destroyed. It can be transferred, or converted, though.

5 0

3 years ago

Read 2 more answers

Draw a mechanism for this reaction. 5-hydroxypentanoic acid forms 2-oxanone in the presence of acid. draw all missing reactants

Licemer1 [7]

Answer: -

The first step involves protonation of the carbonyl oxygen.

After protonation, the Alcohol oxygen now attacks the carbon of the carbonyl.

Thus a six membered ring is formed with 5 carbon atoms and 1 oxygen atom. The 1st position carbon atom has 2 OH groups.

One of these gets again protonated.

This leaves as water. With the loss of the H+, there results a carbonyl at 1 position.

Thus 5-hydroxypentanoic acid forms a lactone or 2-oxanone in presence of acid.

3 0

3 years ago