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sergeinik [125]
3 years ago
15

Which of the following was originally a tenet of Dalton's atomic theory, but had to be revised about a century ago

Chemistry
2 answers:
IgorC [24]3 years ago
4 0
The answer is C. Atoms are <span>tiny indestructible particles.

Hope this helps, feel free to ask any other questions :)</span>
Misha Larkins [42]3 years ago
3 0

If your in Connections Academy these would be the atoms and elements unit test:

1) c.

2) a.

3) b.

4) d.

5) a.

6) c.

7) a.

8) b.

9) b.

10) d.

11) d.

12) c.

13) c.

14) c.

15) a.

16) d.

17) c.

18) c.

19) c.

20) d.

21) c.

22) d.

23) 6 protons, 6 electrons and 7 neutrons

The rest your on your own. Hope this helps ;)

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I have been stuck on this for a while I know you work it out through Hess's law but I can't seem to have the equations balanced
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Answer:

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7 0
2 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

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3 years ago
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Answer:

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Explanation:

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