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3 years ago

Which of the following is an example of kinetic energy?

2 answers:

8 0

Kinetic energy:

SCREAM! -- thundering down a roller coaster path
CRACK -- the baseball flying through the air
Whoosh... -- a crippled plane gliding its way to the airport

Alenkasestr [34]3 years ago

5 0

Something moving. Change in temperature.

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What is the variable for the equilibrium constant Re K Ec E

Nitella [24]

Answer: I am pretty sure it is E

Explanation:

6 0

3 years ago

Why is the following molecule nonpolar and hydrophobic?

algol13

Explanation:

Hydrocarbon shows nonpolar

8 0

3 years ago

A 25.0 ml sample of 0.150 m hydrazoic acid is titrated with a 0.150 m naoh solution. what is the ph after 15.0 ml of the sodium

babunello [35]

First, we need to calculate moles of hydrazoic acid NH3:

moles NH3 = molarity * volume

= 0.15 m * 0.025 L

= 0.00375 moles

moles NaOH = molarity * volume

= 0.15 m * 0.015 L

= 0.00225 moles

after that we shoul get the total volume = 0.025L + 0.015L

= 0.04 L

So we can get the concentration of NH3 & NaOH by:

∴[NH3] = moles NH3 / total volume

= 0.00375 moles / 0.04 L

= 0.09375 M

∴[NaOH] = moles NaOH / total volume

= 0.00225 moles / 0.04 L

= 0.05625 M

then, when we have the value of Ka of NH3 so we can get the Pka value from:

Pka = -㏒Ka

= - ㏒ 1.9 x10^-5

= 4.7

finally, by using H-H equation we can get PH:

PH = Pka + ㏒[salt/ basic]

PH = 4.7 +㏒[0.05625/0.09375]

∴ PH = 4.48

6 0

3 years ago

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N

Lera25 [3.4K]

Answer:

NH3

Explanation:

2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l)

So for two moles of NH3 we need one mole of CO2. So let's count moles for each reagent.

n(NH3)=m(NH3)/M(NH3)=135700/17,03=7968.29 mol

n(CO2)=m(CO2)/M(CO2)=211400/44.01=4803.45 mol

From equation we have to divide n(NH3) by 2 because we need two equivalent per one CO2. That will be 3984.145. So the limiting agent is NH3 because it's not enough of it to react with all CO2

4 0

3 years ago

A solution may contain Ag+, Pb2+, and/or Hg22+. A white precipitate forms when 6 M HCl is added. The precipitate is partially so

omeli [17]

Answer:

All three are present

Explanation:

Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: $AgCl (s), PbCl_2 (s), Hg_2Cl_2 (s)$ .

Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
Secondly, addition of liquid ammonia would form a precipitate with silver: $AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq)$ ; Silver hydroxide at higher temperatures decomposes into black silver oxide: $2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g)$ .
Thirdly, we also know we have $Hg_2^{2+}$ in the mixture, since addition of potassium chromate produces a yellow precipitate: $Hg_2^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow Hg_2CrO_4 (s)$ . The latter precipitate is yellow.

3 0

3 years ago