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Pani-rosa [81]
3 years ago
5

The specific heat of mercury is 380 j/kg/k. How much energy is required to raise the temperature of 42g of mercury from 25°C to

39°C?
please show work and or explain
Thanks in advance!! ​
Chemistry
1 answer:
erma4kov [3.2K]3 years ago
7 0

Answer:

Q = m c T

c= 0.140 j/(g x °c)

m= 250.0g

T =52

hope you can solve it now

Explanation:

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Water that has condensed and formed water droplets on the ground, grass, and other outdoor objects is known as _____.
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A hydrogen halide diffuses 1.49 times faster than HBr. This hydrogen halide is
marysya [2.9K]

To solve this problem, we must assume ideal gas behaviour so that we can use Graham’s law:

vA / vB = sqrt (MW_B / MW_A)

where,

<span>vA = speed of diffusion of A  (HBR)</span>

vB = speed of diffusion of B (unknown)

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MW_A = molar weight of HBr = 80.91 amu

 

We know from the given that:

vA / vB = 1 / 1.49

 

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1/1.49 = sqrt (MW_B / 80.91)

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Since this unknown is also hydrogen halide, therefore this must be in the form of HX.

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6 0
3 years ago
2. Suppose that 21.37 mL of NaOH is needed to titrate 10.00 mL of 0.1450 M H2SO4 solution.
AysviL [449]

Answer:

0.1357 M

Explanation:

(a) The balanced reaction is shown below as:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O

(b) Moles of H_2SO_4 can be calculated as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For H_2SO_4 :

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Volume = 10.00 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 10×10⁻³ L

Thus, moles of H_2SO_4 :

Moles=0.1450 \times {10\times 10^{-3}}\ moles

Moles of H_2SO_4  = 0.00145 moles

From the reaction,

1 mole of H_2SO_4 react with 2 moles of NaOH

0.00145 mole of H_2SO_4 react with 2*0.00145 mole of NaOH

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7 0
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