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olga_2 [115]
3 years ago
10

Determine the Concentration of the Unknown Strong Acid

Chemistry
1 answer:
Natasha2012 [34]3 years ago
3 0
Hey baby I know it’s late but I wanted to say how are u doing
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Which best describes the purpose of this blog post?
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4 years ago
A hot lump of 39.9 g of iron at an initial temperature of 78.1 °C is placed in 50.0 mL H 2 O initially at 25.0 °C and allowed to
Drupady [299]

Answer : The final temperature of the mixture is 29.6^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron = 0.499J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of iron = 39.9 g

m_2 = mass of water  = Density\times Volume=1g/mL\times 50.0mL=50.0g

T_f = final temperature of mixture = ?

T_1 = initial temperature of iron = 78.1^oC

T_2 = initial temperature of water = 25.0^oC

Now put all the given values in the above formula, we get

(39.9g)\times (0.499J/g^oC)\times (T_f-78.1)^oC=-(50.0g)\times 4.18J/g^oC\times (T_f-25.0)^oC

T_f=29.6^oC

Therefore, the final temperature of the mixture is 29.6^oC

8 0
3 years ago
One mole of an ideal gas with a volume of 1.0 L and a pressure of 5.0 atm is allowed to expand isothermally into an evacuated bu
Deffense [45]

Answer:

w= - 1.7173 kJ, q= 1.7173 kJ, q(rev) = 1717.3 J = 1.7173 kJ.

Explanation:

Okay, from the question we are given the information below;

Number of moles, n= 1 mole; initial volume, v(1) = 1.0 litres (L); pressure (p) = 5atm, final volume(v2) = 2.0 Litres(L) ; the workdone, w= not given; the heat, q and q(rev)= not given and the gas was said to expand isothermally.

So, this question is a question from the part of chemistry known as thermodynamics. Therefore, grip yourself we are delving into thermodynamics 'waters' now.

For expansion isothermally; the workdone, w= -nRT ln v2/v1.

Where T= temperature= 25° C = 298 k and R= gas constant.

Therefore; workdone, w = - 1 × 8.314 × 298 × ln(2/1).

Workdone,w= - 1717.32204643. =

- 1717.3 Joules (J).

==> Workdone,w= - 1.7173 kJ.

Then, we are to find q. q can be solved by using the first law of thermodynamics, which by mathematical representation is:

∆U= q + w. Where ∆U= change in internal enegy. Since the question is dealing with isothermal expansion, there is this rule that says for an isothermal expansion ∆U = 0.

Hence, 0 =q + [- 1717.3 Joules (J)].

q=1717.3 J = 1.7173 kJ.

Finally, the q(rev) which is= nRT ln (v2/V1).

q(rev) = 1 × 8.314 × 298 ln (2/1).

q(rev) = 1717.3 J = 1.7173 kJ.

PS: please note the negative signs in the workdone and the positive sign in the q(rev).

7 0
3 years ago
Which of the following terms means that metals can be pulled into thin strands or wires? [like copper]
Dimas [21]

Answer:

D

Explanation:

plz mark brainliest answer if it helps

5 0
3 years ago
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