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4 years ago

Which parts (postulates) of Dalton's atomic theory has been modified in light of later discoveries

Chemistry

1 answer:

professor190 [17]4 years ago

8 0

Explanation:

1) Atoms can not be subdivided. It has been changed as it is possible to divide an atom into Protons, Neutrons and Electrons as well as other smaller particles.

2) The regulations for the chemical mixture have always been altered to the rules for the chemical combination in the creation of organic compounds may be denied.

3)Atoms are of a particular element are identical in all respects i. e. they have same mass and similar properties.

It was modified. And according. to new search;

Atoms of the same element may not be always identical.

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3 years ago

PLEASE HELP! DUE TODAY! How many gram of magnesium react with 120 ml H3PO4 (with molar concentration 2M)?

nika2105 [10]

Answer:

3.89 g Mg

Explanation:

2M = 2 mol/L

120 mL = 0.120 L

2 mol/L *0.120 L = 0.240 mol H3PO4

3Mg + 2H3PO4 ----> Mg3(PO4)2 + 3H2

from reaction 3 mol 2 mol

given 0.240mol x mol

x = 0.240*2/3 = 0.16 mol Mg

M(Mg) = 24.3 g/mol

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3 years ago

What would happen if you didn't have chemical energy in your body

astra-53 [7]

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3 years ago

How many grams of magnesium are needed to completely react

Verizon [17]

97.22 grams of magnesium is needed.

Explanation:

The molar ratio of Mg to O2 to make MgO is 2:1

Here, we have 2 moles of O2, therefore we need 4 moles of Mg.

1 mol of Mg = 24.305 g

Therefore

4 moles of Mg = 4 $\times$ 24.305 g

= 97.22 g

Hence we need 97.22grams of magnesium are needed to completely reactwith 2.00 mol of O2 in the synthesis reaction that producesmagnesium oxide.

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3 years ago

A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the reaction between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now for example, we can use a volume of 10 mL of HCl. So, we can calculate the moles using the molarity equation:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a concentration, in this case KOH is in excess, so we have to calculate the concentration of KOH. For this, we already have the moles of KOH that remains left, now we need the total volume:

$Total~volume=50mL+10mL=60mL$