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Slav-nsk [51]
3 years ago
7

The joule is an SI unit of energy that is equal to 1 kg•m2/s2. The erg is a different unit of energy that is equal to 1 g•cm2/s2

. How many ergs are in one joule? Use dimensional analysis to solve this problem.
Chemistry
1 answer:
vredina [299]3 years ago
7 0

Answer:

The answer to your question is: 1 J = 10, 000, 000 ergs

Explanation:

(1 kgm²/s²)(1000 g/ 1 kg) = 1000 gm²/s²

(1000 gm²/s²)(100 cm/1 m) = 1 00 000 gmcm/s²

1 00 000 gmcm/s² (100 cm / 1m) = 10 000 000 gcm²/s²

                                                       = 10 000 000 ergs

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How many moles of potassium hydroxide are needed to completely react with 2.94 moles of aluminum sulfate according to the follow
fiasKO [112]

Answer:- Third choice is correct, 17.6 moles

Solution:- The given balanced equation is:

Al_2(SO_4)_3+6KOH\rightarrow 2Al(OH)_3+3K_2SO_4

We are asked to calculate the moles of potassium hydroxide needed to completely react with 2.94 moles of aluminium sulfate.

From the balanced equation, there is 1:6 mol ratio between aluminium sulfate and potassium hydroxide.

It is a simple mole to mole conversion problem. We solve it using dimensional set up as:

2.94molAl_2(SO_4)_3(\frac{6molKOH}{1molAl_2(SO_4)_3})

= 17.6 mol KOH

So, Third choice is correct, 17.6 moles of potassium hydroxide are required to react with 2.94 moles of aluminium sulfate.


8 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
4 years ago
Which one of the following
mariarad [96]

Answer:

5

Explanation:

6 0
3 years ago
Calculate the molarity of sodium chloride in a half-normal saline solution (0.45% NaCl). The molar mass of NaCl
Cerrena [4.2K]

Answer:

0.077 M

Explanation:

Molarity is the representation of the solution.

Molarity:

It is amount of solute in moles per liter of solution and represented by M

Formula used for Molarity

                    M = moles of solute / Liter of solution . . . . . . . . . . (1)

Data Given :

The concentration of half normal (NaCl) saline = 0.45g / 100 g

So,

Volume of Solution =  100 g = 100 mL

Volume of Solution in L =  100 mL / 1000

Volume of Solution =  0.1 L

molar mass of NaCl = 58.44 g/mol

Now to find number of moles of Nacl

               no. of moles of NaCl = mass of NaCl / molar mass

               no. of moles of NaCl =  0.45g / 58.44 g/mol

               no. of moles of NaCl =  0.0077 g

Put values in the eq (1)

                  M = moles of solute / Liter of solution . . . . . . . . . . (1)

                  M = 0.0077 g / 0.1 L

                  M = 0.077 M

So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M

3 0
3 years ago
Which is a cause of physical weathering but not of chemical weathering ??
Soloha48 [4]
Wind becuase you can feel it whhich is physical 

8 0
4 years ago
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