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2 years ago

It's urgent! Any kind soul please help me ASAP!

Chemistry

1 answer:

nika2105 [10]2 years ago

8 0

Answer:

a white precipitate of barium chloride is formed as displacement reaction takes place.

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How many atoms are in Sodium, Carbon, Hydrogen, Oxygen, Fluorine, Boron, Lithium, Helium, Phosphorus and Sulfur?

vlada-n [284]

Sodium. 11
Carbon. 12
Hydrogen 1
Oxygen 2
Fluuorine. 14
Boron. 5
Lithium. 6
Helium 3
Phosphorus 15
Sulfur 6

6 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

Find the number of moles in 3.62 x 1024 atoms of helium.

Makovka662 [10]

3.62x10^24/ 6.02x10^23= 6.013 moles to 3dp

8 0

3 years ago

Read 2 more answers

Which lists only metalloids?

enot [183]

Answer:

boron (B), germanium (Ge), and tellurium (Te)

6 0

2 years ago

Read 2 more answers

A 110 g copper bowl contains 240 g of water, both at 21.0°C. A very hot 410 g copper cylinder is dropped into the water, causing

vlada-n [284]

Answer:

There is 98.76 kJ energy transfered to the water as heat.

Explanation:

Step 1: Data given

Mass of copper bowl = 110 grams

Mass of water = 240 grams

Temperature of water and copper = 21.0 °C

Mass of the hot copper cylinder = 410 grams

8.6 grams being converted to steam

Final temperature = 100 °C

Step 2: Calculate the energy gained by the water:

Q = m(water)*C(water)*ΔT + m(vapor)*Lw

⇒with mass of water = 0.240 kg

⇒ with C(water) = the heat capacity of water = 4184 J/kg°C

⇒ with ΔT = the change in temperature = T2 - T1 = 100 °C - 21.0 = 79°C

⇒ with mass of vapor = 8.60 grams = 0.0086 kg

⇒ with Lw = The latent heat of vaporization (water to steam) = 22.6 *10^5 J/kg

Q = 0.24kg * 4184 J/kg°C *79°C + 0.0086 kg*22.6*10^5 J/kg

Q = 79328.64 + 19436 = 98764.64 J = 98.76 kJ

There is 98.76 kJ energy transfered to the water as heat.

4 0

3 years ago

It's urgent! Any kind soul please help me ASAP! ​

It's urgent! Any kind soul please help me ASAP!