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3 years ago

3. A sample of carbon atoms were chemically combined with a sample of oxygen atoms to yield the

1 answer:

5 0

Your mom is the answer

Explanation: moms tend to be the only formula to wake you up for school when no one can
1. Gives you a heart attack
2. Cusses at you
3. You wake up

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How can a Hydrogen atom turn into a<br> Hydrogen ion (H+)?

Butoxors [25]

Answer:

By losing an electron

Explanation:

Electrons have a negative charge. So, losing one would give an element a more positive charge. You can usually find a hydrogen ion (H+) in substances like acids.

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3 years ago

What are 3 permeable materials

gizmo_the_mogwai [7]

Permeable materials for what

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3 years ago

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Converting carbon to nitrogen by increasing the number of protons from 6 to 7 and decreasing the number of neutrons from 8 to 7

Anton [14]

this is beta decay as the mass number stays the same but proton number changes, this is specifically beta minus as a neuron changes into a proton

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3 years ago

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Writing the net ionic equation

vazorg [7]

First write the molecular equation with states:

(NH4)2S (aq) + 2AgNO3(aq) → Ag2S (s) + 2NH4NO3

Now write a full ionic equation by separating into ions all substances that dissociate: anything (s) (g) or (l) does not dissociate

2NH4 + (aq) + S 2-(aq) + 2Ag+ (aq) + 2NO3- (aq) → Ag2S(s) + 2NH4 + (aq) + 2NO3- (aq)

To write the NET IONIC equation, inspect the full ionic equation above and delete anything that appears on both sides of the → sign:

Net ionic equation:

S 2-(aq) + 2Ag + (aq) → Ag2S(s)

3 0

3 years ago

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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