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Neporo4naja [7]
3 years ago
10

60 х 8 у x= 04 09 O 16

Chemistry
1 answer:
vovikov84 [41]3 years ago
7 0

Answer:

The answer is 16

Explanation:

It needs to be hald of the medium so devide the 2 numbers

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Select the correct electron configuration for Boron. (Atomic Number 5) 1s 22s 12p 2 1s 22s 3 1s 32s 12p 1 1s 22s 22p 1
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ans is: 1s2, 2s2 ,2p1,...

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3 years ago
Please answer, with explanation. Thanks!​
nadya68 [22]

Answer:

Explanation:

a = 40.1 g of Ca

Number of moles = mass / molar mass

Number of moles = 40.1 g/ 40.1 g/mol

Number of moles = 1 mol

b = 11.5 g Na

Number of moles = mass / molar mass

Number of moles = 11.5 g/ 23 g/mol

Number of moles = 0.5 mol

c = 5.87 g Ni

Number of moles = mass / molar mass

Number of moles = 5.87 g/ 58.7 g/mol

Number of moles = 0.1 mol

d = 150 g of S

Number of moles = mass / molar mass

Number of moles = 150 g/ 32 g/mol

Number of moles = 4.7 mol

e = 2.65 g Fe

Number of moles = mass / molar mass

Number of moles = 2.65 g/ 55.85 g/mol

Number of moles = 0.05 mol

f = 0.00750 g Ag

Number of moles = mass / molar mass

Number of moles = 0.00750 g/ 107.9 g/mol

Number of moles = 6.95 × 10⁻⁵ mol

g = 2.25 × 10²⁵ atoms Zn

1 mole = 6.022 × 10²³ atoms

1 mol / 6.022 × 10²³ atoms × 2.25 × 10²⁵ atoms

0.17  × 2.25 × 10²⁵ moles

38.25 moles

h = 50 atoms of Ba

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1 mol / 6.022 × 10²³ atoms ×50 atoms

0.17 × 10²³ × 50 moles

8.5 × 10²³ moles

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3 years ago
For each pair, predict which molecule has the greater molar entropy under the same conditions (assume gaseous species).c o 2, ca
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8 0
3 years ago
Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find
FromTheMoon [43]

<u>Answer:</u> The concentration of SO_4^{2-} at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

       H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)

            0.025          0.025       0.025

Equation for the second dissociation of sulfuric acid:

                    HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)

<u>Initial:</u>            0.025            0.025      

<u>At eqllm:</u>      0.025-x          0.025+x        x

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

We know that:

Ka_2\text{ for }H_2SO_4=0.01

Putting values in above equation, we get:

0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of SO_4^{2-} at equilibrium is 0.00608 M

4 0
3 years ago
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