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worty [1.4K]
2 years ago
15

Realizar un Experimento Teóricamente, considerando la definición de mezcla, dibuja y explicar de forma teórica su proceso de pre

paración de una mezcla de su preferencia y su elección​
Chemistry
1 answer:
Mrrafil [7]2 years ago
7 0

Answer:

rffdf

Explanation:

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BRAINLIESTTT ASAP!!! PLEASE HELP ME :)
Leno4ka [110]

Answer:

"Thermometer C, because it measures accurately in the ones place."

Explanation:

Thermometer D measures using tens place. Since we are measuring the liquid's temperature at 47 degrees Celsius, the most appropriate thermometer would measure in ones place, not tens place.

Hope this helps! :)

8 0
3 years ago
Which of the following lists only include molecules? He h2o
lbvjy [14]
He is the only one that includes only mollecules
4 0
3 years ago
Which of these is the part of an atom that contains the protons and neutrons? A. nucleus B. electron C. shell D. orbital
Ludmilka [50]
The answer is a, nucleus
4 0
3 years ago
Read 2 more answers
How much water can be raised from 25*C (room temperature) to 37*C (body temperature) by adding the 2,000 kJ in a Snickers bar?
yuradex [85]

Answer:

m = 39834.3 g

Explanation:

Given data:

Mass of water raised = ?

Initial temperature = 25°C

Final temperature = 37°C

Energy added = 2000 Kj (2000 ×1000= 2000,000 j

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT =  37°C - 25°C

ΔT = 12°C

c = 4.184 g/j.°C

Q = m.c. ΔT

2000,000j = m .4.184 g/j.°C. 12°C

2000,000j = m. 50.208 g/j

m =  2000,000j / 50.208 g/j

m = 39834.3 g

6 0
2 years ago
N2 + 3H2 Right arrow. 2NH3 What is the percent yield of NH3 if the reaction of 26.3 g of H2 produces 79.0 g of NH3? Use Percent
Romashka [77]

Answer:

53.4%

Explanation:

N₂ + 3H₂ → 2NH₃

Given that:

26.3 g of H₂ react with N₂ to produce 79.0 g of NH₃

Then:

The number of moles of H₂ = 26.3g of H₂ * (1 mol of H₂/ 2.02g of H₂)

= 13.05 mol of H₂

The number of moles of NH₃ = 13.05 mol of H₂ * ( 2mol of NH₃/ 3 mol of H₂)

= 8.697 mol of NH₃

The mass of NH₃ = 8.697 mol of NH₃ *( 17.04g of NH₃/ 1 mol of NH₃) = 148.1 g of NH₃

The percent yield = actual yield/ theoretical yield * 100%

The percent yield = ( 79.0 g/ 148.1 g )* 100%

The percent yield ≅ 53.4 %

7 0
2 years ago
Read 2 more answers
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