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alekssr [168]
3 years ago
10

Elemento quimico con inicial "S"​

Chemistry
1 answer:
gtnhenbr [62]3 years ago
4 0

Answer:

El elemento es azufre.

Explanation:

Espero que esto ayude a marcar el MÁS CEREBRAL !!!

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The melting point of unknown substance A is 15°C and its boiling point is 95°C. Sketch a
Igoryamba

Answer:

Idk if this is right but i hope it helps... sorry if it's wrong

Explanation:

4 0
3 years ago
A beaker containing 6.32 moles of PBr3, contains___
NISA [10]

Answer:

3.8 x 10²⁴molecules

Explanation:

Given parameters:

Number of moles  = 6.32moles

Unknown:

Number of molecules  = ?

Solution:

The number of moles can be used to derive the number of molecules found within a substance.

Now,

       1 mole of substance contains 6.02 x 10²³ molecules

     6.32 mole of PBr₃ will contain 6.32 x 6.02 x 10²³ = 3.8 x 10²⁴molecules

8 0
3 years ago
By titration, it is found that 28.5 mL of 0.183 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the
maksim [4K]

Answer:

0.209M

Explanation:

M1V1=M2V2

(28.5 mL)(0.183M)=(25.0mL)(M)

M2= 0.209M

*Text me at 561-400-5105 for private tutoring if interested: I can do homework, labs, and other assignments :)

4 0
3 years ago
How much heat is evolved in converting 1.00 mol of steam at 155.0 ∘c to ice at -50.0 ∘c? the heat capacity of steam is 2.01 j/(g
Ne4ueva [31]
When the specific heat capacity of the water is 4.18 J/g.°C so, we are going to use this formula to get the heat for cooling three  phases changes from steam to liquid and from liquid to ice (solid) :

when Q = M*C*ΔT 

Q is the heat in J

and M is the mass in gram = 1 mol H2O * 18 g/mol(molar mass) = 18 g

C is the specific heat J/g.°C

ΔT is the change in temperature

Q = Mw *[ ( Csteam * ΔTsteam)+(Cw*ΔTw) + (Cice  * ΔT ice)]

    = 18 g * [(2.01 * (155-100°C)) + (4.18 * (100-0°C)) + (2.09 * (0 - 55 °C))]

∴Q = 7444.8 J

and when we know that the heat of fusion for water = 334J/g

and heat of vaporization for water =  2260J/g


∴Q for the two phases changes = M * (2260+334) 

                                                      = 18 * (2260+334)

                                                      = 46692 J 

∴ Q total = 7444.8 + 46692 = 54136.8 J
5 0
3 years ago
If the titrant has a molarity of 0.1000 m and there are 45.00 ml of analyte present, what is the molarity of the analyte?
marysya [2.9K]
Given: 

Concentration of titrant = 0.1000 M
Volume of titrant = 45 mL

The molarity of analyte depends on the amount of the analyte present in the titrated solution. If the amount of analyte is 20 mL, then its concentration is:

45ml * 0.10 M = C analyte * 20 ml

C analyte = 0.225 M
4 0
3 years ago
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