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3 years ago

What is the relative molecular mass (mr) of ammonia, formula NH3?

Chemistry

1 answer:

egoroff_w [7]3 years ago

7 0

Answer:

APROXIMENTLY 17

Explanation:

lol i remebered from science class

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I have little or no rainfall for long periods of time. I cause death to all living things because of lack of water. What am I?

Doss [256]

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3 years ago

Hydrogen bond is an example for? (PLEASE HELP!)

zlopas [31]

Answer:

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3 years ago

375 mL of a 0.88 M potassium hydroxide solution is added to 496 mL of a 0.76 M cesium hydroxide solution. Calculate the pOH of t

AysviL [449]

Answer:

pOH of resulting solution is 0.086

Explanation:

KOH and CsOH are monoacidic strong base

Number of moles of $OH^{-}$ in 375 mL of 0.88 M of KOH = $\frac{0.88\times 375}{1000}moles$ = 0.33 moles

Number of moles of $OH^{-}$ in 496 mL of 0.76 M of CsOH = $\frac{0.76\times 496}{1000}moles$ = 0.38 moles

Total volume of mixture = (375 + 496) mL = 871 mL

Total number of moles of $OH^{-}$ in mixture = (0.33 + 0.38) moles = 0.71 moles

So, concentration of $OH^{-}$ in mixture, $[OH^{-}]$ = $\frac{0.71}{871}\times 1000M=0.82M$

Hence, $pOH=-log[OH^{-}]=-log(0.82)=0.086$

8 0

3 years ago

Which of the following substances found in tobacco smoke stimulates the brain? A. acetone B. toluene C. nicotine D. ammonia Plea

Vaselesa [24]

The answer is C. Nicotine is the substance found in tobacco smoke that stimulates the brain.

7 0

3 years ago

Read 2 more answers

Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr

VikaD [51]

Answer : The value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

Explanation :

The balanced cell reaction will be,

$Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)$

The half-cell reactions are:

Oxidation reaction (anode) : $Pb(s)\rightarrow Pb^{2+}(aq)+2e^-$

Reduction reaction (cathode) : $2Ag^+(aq)+2e^-\rightarrow 2Ag(g)$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell = 0.93 V

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

$\Delta G^o=-2\times 96500\times 0.93$

$\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol$

Now we have to calculate the value of 'K'.

$\Delta G^o=-RT\ln K$

where,

$\Delta G_^o$ = standard Gibbs free energy = -180 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

$-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K$

$K=3.6\times 10^{31}$

Therefore, the value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

5 0

3 years ago