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liq [111]
3 years ago
12

How much energy is required to heat 14 grams of water from 18 degrees Celsius to 47 degrees Celsius

Chemistry
1 answer:
Alborosie3 years ago
5 0

Answer:

Q=1,698.7J

Explanation:

Hello!

In this case, since the calculation of the required energy is performed via the following equation:

Q=mC(T_f-T_i)

Thus, since the mass is 14 g, specific heat is 4.184 J(g*°C) and the temperatures are 47 °C and 18 °C respectively, the resulting energy is:

Q=14g*4.184\frac{J}{g\°C}(47°C-18°C)\\\\Q=1,698.7J

Best regards!

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Answer:

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8 0
3 years ago
1. How many grams would 8.1 x 1021 molecules of sucrose (C12H22011)<br> weigh?
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Answer:

Mass = 4.6 g

Explanation:

Given data:

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Mass of sucrose = ?

Solution:

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1 mole × 8.1 ×10²¹ molecules / 6.022×10²³ molecules

1.35 × 10⁻² mol

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7 0
4 years ago
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Ilia_Sergeevich [38]

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8 0
3 years ago
Read 2 more answers
If all other variables remain unchanged, what happens to the output force when the area of the input piston is doubled?
mars1129 [50]

Answer:

1250N

Explanation:

This question is based on pascal's Law.

So By Pascal's Law

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                  = Force or weight on output person.

therefore after putting the values we get,

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3 0
3 years ago
What is the temperature of a gas that is expanded from 3.75 L at 37 degrees Celsius to 5.6 L?
deff fn [24]

Answer:

190 °C  

Step-by-step explanation:

The pressure is constant, so this looks like a case where we can use <em>Charles’ Law</em>:  

V₁/T₁ = V₂/T₂      Invert both sides of the equation.  

T₁/V₁ = T₂/V₂      Multiply each side by V₂

T₂ = T₁ × V₂/V₁

=====

V₁ = 3.75 L; T₁ = (37 + 273.15) K = 310.15 K  

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=====

T₂ = 310.15 × 5.6/3.75

T₂ = 310.15 × 1.49

T₂ = 463 K

t₂ = 463 – 273.15

t₂ = 190 °C

3 0
3 years ago
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