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3 years ago

Round the nearseat whole number how many neutrons on average are in atom of osmium?

Chemistry

1 answer:

Anettt [7]3 years ago

7 0

Answer:

jdjdjdhjxjxjchxjxjkxjxjxhxnxkjcjcjcj

Explanation:

sorry i just need point lolllllllllll

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A force of 3000N is applied on a 20kg ball. Find the acceleration of the ball

Andrews [41]

Answer:

force = 3000N

mass= 20 kg

now

F= ma

3000= 20×a

3000÷20=a

a=15

F= ma by newtons second law of motion

3 0

2 years ago

Question 10

olga2289 [7]

Answer:

Here it would be A,B and C because all the examples are violating safety precautions. They are all examples of un safe signs.

Hope this helps

Explanation:

6 0

3 years ago

The solubility product of AgCl is 1.4 x 10-4 at 100°C. Calculate the solubility of AgCl in

marshall27 [118]

Answer:

AgCl=Ag+ + Cl-

1.4×10^-4 =x × x

x²=1.4×10^-4

x=√(1.4×10^-4)

x=0.012

5 0

3 years ago

Which of the following possess the greatest concentration of hydroxide ions?

jek_recluse [69]

Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

(a) a solution of pH 3.0

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$11=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-11}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-11}M$

(b) a solution of 0.10 M $NH_3$

As we know that 1 mole of $NH_3$ is a weak base. So, in a solution it will not dissociates completely.

So, the $OH^-$ concentration will be less than 0.10 M

(c) a solution with a pOH of 12.

We have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$12=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-12}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-12}M$

(d) a solution of 0.10 M NaOH

As we know that $NaOH$ is a strong base. So, it dissociates to give $Na^+$ ion and $OH^-$ ion.

So, 0.10 M of $NaOH$ in a solution dissociates to give 0.10 M of $Na^+$ ion and 0.10 M of $OH^-$ ion.

Thus, the $OH^-$ concentration is, 0.10 M

(e) a $1\times 10^{-4}M$ solution of $HNO_2$

As we know that 1 mole of $HNO_2$ in a solution dissociates to give 1 mole of $H^+$ ion and 1 mole of $NO_2^-$ ion.

So, $1\times 10^{-4}M$ of $HNO_2$ in a solution dissociates to give $1\times 10^{-4}M$ of $H^+$ ion and $1\times 10^{-4}M$ of $NO_2^-$ ion.

The concentration of $H^+$ ion is $1\times 10^{-4}M$

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (1.0\times 10^{-4})$

$pH=4$

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$10=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-10}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-10}M$

From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.

Hence, the correct option is (d)

3 0

3 years ago

How is the electron cloud model different from Bohr's atomic model?

Cerrena [4.2K]

In the electron cloud model, the atoms are in unpredicted places. But in the Bohr model, atoms are in "rows"

7 0

3 years ago