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3 years ago

Help asapp plzz!!!!

Chemistry

1 answer:

tino4ka555 [31]3 years ago

7 0

Answer:

Determining what data to collect.

Explanation:

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Find the density of an object that has a mass of 5 kg and a volume of 50 cm3.

iren2701 [21]

Mass=5kg=5000g
Volume=50cm^3

$\\ \rm\longmapsto Density=\dfrac{Mass}{Volume}$

$\\ \rm\longmapsto Density=\dfrac{5000}{50}$

$\\ \rm\longmapsto Density=100g/cm^3$

4 0

3 years ago

Which is not considered a water flow?

Artemon [7]

Rain maybe?? Cause it doesn’t flow??? It comes out of the sky?? Sorry if its wrong or sth

6 0

3 years ago

If u mix 10ml of 70%ethanol with 20ml of 80% ethanol with 50ml water what is the concentration of the resulted solution

morpeh [17]

Answer:

29% is the final concentration

Explanation:

To solve this question we must find the volume of ethanol added and the volume of the whole solution. The concentration will be:

Volume ethanol / Total volume * 100

Volume ethanol:

10mL * 70% = 7mL ethanol

20mL * 80% = 16mL ethanol

= 23mL ethanol

Total volume:

10mL + 20mL + 50mL = 80mL

Concentration:

23mL / 80mL * 100

= 29% is the final concentration

6 0

3 years ago

Mixing hydrochloric acid (HCl) and aqueous calcium hydroxide produces water and aqueous Calcium Chloride. Write the complete and

swat32

Answer:

HCI+Ca(OH)2

Explanation:

6 0

3 years ago

Consider the insoluble compound silver bromide , AgBr . The silver ion also forms a complex with ammonia . Write a balanced net

Degger [83]

Answer:

- $AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)$

- $K=1.2x10^{-5}$

Explanation:

Hello,

In this case, by considering the dissolution of silver bromide:

$AgBr(s)\rightleftharpoons Ag^+(aq)+Br^-(aq) \ \ \ Ksp=[Ag^+][Br^-]=7.7x10^{-13}$

And the formation of the complex:

$Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)\ \ \ Kf=\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}=1.6x10^7$

We obtain the balanced net ionic equation by adding the aforementioned equations:

$AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)+Ag^+(aq)\\\\AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)$

Now, the equilibrium constant is obtained by writing the law of mass action for the non-simplified net ionic equation:

$AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-+Ag^+\\\\K=[Ag^+][Br^-]*\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}$

So we notice that the equilibrium constant contains the solubility constant and formation constant for the initial reactions:

$K=Ksp*Kf=7.7x10^{-13}*1.6x10^{7}\\\\K=1.2x10^{-5}$

Best regards.

4 0

3 years ago