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Dafna11 [192]
3 years ago
5

The Element rhenium has two naturally occurring isotopes, 185 Re and 187 Re, with an average atomic mass of 186.207 amu. Rhenium

is 62.60% 187 Re, and the atomic mass of 187 Re is 186.956 amu. Calculate the mass of 185 Re. How do you do this?
Chemistry
1 answer:
BigorU [14]3 years ago
8 0
To calculate the average mass of the element, we take the summation of the product of the isotope and the percent abundance. In this case, the equation becomes 186.207=187*0.626+185*x where x is the percent abundance of 185. The answer is 0.374 or 37.4%. This can also be obtained by 100%-62.6%= 37.4%. 
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When 4.41g of phosphoric acid (H3PO4) react with 9.25g of barium hydroxide, water and insoluble barium phosphate form. [T/I-7] a
AnnZ [28]

Answer:

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

Explanation:

Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.

H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

We will balance it using the trial and error method.

First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

Finally, we will get the balanced equation by multiplying H₂O by 6.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

3 0
3 years ago
What would be the mass of 9.03*10^21 molecules of hydrobromic acid
Fynjy0 [20]

Answer:

2.11 g hydrobromic acid (correct to 3SF)

Explanation:

Molecular formula of hydrobromic acid = C2H5BrO2

mass of C2H5BrO2 = 140.96g

Beginning with what we're given, 9.03*10^21 we then make a conversion by using Avegadro's number which is 6.02*10^23 per mole (Oct. 23 at 6:02 am is national mole day :) Then, we need to convert out of moles, 140.96g hydrombromic acid per mole.

It looks like this:

9.03*10^21 molecules • (1 mol C2H5BrO2 / 6.02*10^23 molecules) • (140g C2H5BrO2 / 1 mol) = 2.1144 g C2H5BrO2

3 0
3 years ago
Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang
Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

\Delta (H_{f})_{MnO_{2}} = -520.0 KJ/mole

\Delta (H_{f})_{Al_{2}O_{3}} = -1699.8 KJ/mole

The balanced chemical reaction is,

3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)

Formula used :

\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}

\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]

                        = (-3399.6) + (1560)

                        = -1839.6 KJ



5 0
3 years ago
Which would have a greater density, 5 g of aluminum or 5 kg of aluminum if they were both the same temperature? Fully explain yo
leva [86]

Answer:the 5g would be more dense due to the mass and structure

Of it causing the atoms to expand as heat gets added

Explanation:

8 0
3 years ago
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