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Rina8888 [55]
3 years ago
5

How to calculate percent error of a melting point range?

Chemistry
1 answer:
olya-2409 [2.1K]3 years ago
5 0

Answer:

3.9%

Explanation:

Exact melting point = 123°C

Measured melting point = 128°C

% error; |measured - exact|/exact × 100

% error = |123 -128|/128 × 100

%error = 3.9%

Percentage error in temperature measurement = 3.9 %

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5 0
3 years ago
Read 2 more answers
What mass of a 4.00 % NaOH solution by mass contains 15.0 g of NaOH?
fenix001 [56]

Answer:

375 g

Explanation:

Let the total mass of the solution = x g

Given that the mass of NaOH in the solution = 15.0 g

Thus, mass of the solvent = x - 15.0 g

Given that the mass % of the solution = 4 %

So,

Mass\ percent=\frac {Mass\ of\ the\ solute}{Mass\ of\ the\ solution}\times 100

So,

4=\frac {15.0}{Mass\ of\ the\ solution}\times 100

<u>Mass of the solution = 375 g</u>

7 0
4 years ago
What is δs° for the reaction so2(s) no2(g) → so3(g) no(g)? substance s°(j/k • mol) so2(g) 248.5 so3(g) 256.2 no(g) 210.6 no2(g)
zimovet [89]

The value of ΔS° for reaction  is - 22.2 J/K.mol

SO_{2}(s)+NO_{2} (g) → SO_{3}(g)+NO(g)

Calculation,

Given value of S°(J/K.mol) for

SO_{2}(s) = 248.5

NO_{2} (g) = 240.5

NO(g) = 210.6

SO_{3}(g) = 256.2

Formula used:

ΔS° (Reaction) =  ∑S°(Product) - ∑S°(Reactant)

ΔS° = (256.2 + 210.6 ) - ( 248.5 + 240.5) = 466.8 - 489 = - 22.2 J/K.mol

The change in stander entropy of reaction is  - 22.2 J/K.mol. The negative sign indicates the that entropy of reaction is decreases when reactant converted into product.

learn about reaction

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3 0
2 years ago
What is the charge on an electron?<br> O-1<br> O 0<br> O +1<br> O +2
GarryVolchara [31]

Answer:

-1

Explanation:

Because it is the smallest sub-atomic particle

7 0
3 years ago
A gas in a rigid container at 25°C has a pressure of 0.96 atm. A change in temperature causes the pressure to increase to 1.25 a
zalisa [80]

Answer:

Answer is c....115°C

Explanation:

Gay-Lussacs law says that pressure of a gas is the same as temperature when the volume is kept the same

P / T = k

where P - pressure , T - temperature in kelvin and k - constant

parameters for the first time are on the left side of the equation and parameters for the second instance are on the right side of the equation

T1 - 25 °C + 273 = 298 K

substiuting the values in the equation

T2 = 388 K

temperature in celcius - 388 K - 273 = 115 °C

6 0
3 years ago
Read 2 more answers
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