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Gnom [1K]
3 years ago
15

Asolutioncontains50gofsugarin250g ofwater.Calculatetheconcentrationin termsofmassbymasspercentageofthe solution.​

Chemistry
1 answer:
vladimir1956 [14]3 years ago
8 0
Can you explain this better please thank u
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Al comenzar la reacción: N2(g) + 2O2(g) ------> 2NO2(g) existe 1 mol de N2 y 2 moles de O2 y al
miss Akunina [59]

Answer:

65

Explanation:

estion

Al comenzar la reacción: N2(g) + 2O2(g) ------> 2NO2(g) existe 1 mol de N2 y 2 moles de O2 y al

finalizarla está presente una mezcla formada por 2,2 moles en total, ¿cuál es el rendimiento para la

reacción?

7 0
3 years ago
Calcular la normalidad de 1 Kg de sulfuro de aluminio en 5000 ml de solucion.
Troyanec [42]

Calculate the normality of 1 Kg of aluminum sulfide in 5000 ml of solution.

Normality comes out to be 8.11

<h3> Given </h3>
  • Mass of solute: 1000g
  • Volume of solution (V): 5000 ml = 5 liters
  • Equivalent mass of solute (E) = molar mass / n-factor

n-factor for Al_{2}S_{3} is 6 and molar mass is 148g

So, on calculating equivalent mass is equal to 24.66g

FORMULAE of Normality (N) = (Mass of the solute) / (Equivalent mass of the solute (E) × Volume of the solution (V)

                                           N=\frac{1000}{24.66*5}

                                          <u> N=8.11</u>

Therefore, normality of 1 kg aluminum sulfide is 8.11

Learn more about normality here brainly.com/question/25507216

#SPJ10

7 0
2 years ago
Calculate the number of moles in 200 grams of Sodium hydroxide​
TEA [102]

Answer:

The answer is 5 mol

Explanation:

Number Of Moles = Mass Of Substance ÷ Mass Of One Mole

Mass of NaOH = Molar Mass of Sodium + Oxygen + Hydrogen

23 + 16 + 1

40 g/mol

Number of Mole = 200/40 = 5 mol

Thus, The number of Moles in 200 grams of Sodium hydroxide is 5 mol

<u>-TheUnknownScientist 72</u>

6 0
3 years ago
1,000 millimeters is the same as _____
Aleksandr [31]

Answer:

1 meter

Explanation:

6 0
4 years ago
Read 2 more answers
Which ligand binds the tightest? Ligand A, with a Kd of 10^-9M Ligand B, with a Kd of 10^-3M Ligand C, with a percent occupancy
padilas [110]
To answer this item, we must take note that the ligand that binds the tightest is the one with the lowest dissociation constant, Kd. Kd's for both A and B are already given so, we only need to solve Kds for C and D. 
  Kd of C
                        0.3 = (1x10⁻⁶)/(1x10⁻⁶ + Kd)   ; Kd = 2.3x10⁻⁶
 Kd of D  
                        0.8 = (1x10⁻⁹)/(1x10⁻⁹ + Kd)   ; Kd = 2.5x10⁻10
Since Ligand D has the least value of dissociation constant then, it can be concluded that it binds the tightest. 
6 0
3 years ago
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