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3 years ago

20 mL of 80°C water is mixed with 20 mL of 0°C water in a perfect calorimeter. What is the final temperature?

1 answer:

8 0

To calculate for the final temperature, we need to remember that the heat rejected should be equal to the absorbed by the other system. We calculate as follows:

Q1 = Q2
(mCΔT)1 = (mCΔT)2

We can cancel m assuming the two systems are equal in mass. Also, we cancel C since they are the same system. This leaves us,

(ΔT)1 = (ΔT)2
(T - 80) = (0 - T)
T = 40°C

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Which group of elements are considered to be in the same family?

mash [69]

A. helium, neon and argon, because they are in the same group or column

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3 years ago

An unknown substance is heated from 10 °C to 25 °C. The table below shows the state of the substance at different temperatures.

bezimeni [28]

Answer:

Its melting point is 17 °C.

Explanation:

The melting point of a substance is the temperature at which it changes state from solid to liquid. At the melting point the solid and liquid phase exist in equilibrium.

So, the melting point is 17 °C.

And this is shown in the figure attached.

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3 years ago

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How many grams of P2Cl5 are produced if 2.00 moles of P reacts with an excess of Cl2?

Roman55 [17]

Answer:

$m_{P_2Cl_5 }=239.2125gP_2Cl_5$

Explanation:

Hello there!

In this case, according to the reaction:

$4P+5Cl_2\rightarrow 2P_2Cl_5$

Thus, since there is a 4:2 mole ratio of P to P2Cl5 and the molar mass of the later is 239.2125 g/mol, we obtain the following mass as the produced one:

$m_{P_2Cl_5 }=2.00molP*\frac{2molP_2Cl_5}{4molP} *\frac{239.2125gP_2Cl_5 }{1molP_2Cl_5} \\\\m_{P_2Cl_5 }=239.2125gP_2Cl_5$

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3 years ago

Brainiest if answered in the next 5m

bearhunter [10]

The equation is already balanced

2AgNO3 + MgCl2 => 2AgCl + Mg(NO3)2

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3 years ago

The tarnish that forms on objects made of silver is solid silver sulphide. This can be removed by reacting it with aluminium met

steposvetlana [31]

Answer:

5.83 mol.

Explanation:

From the balanced reaction:

2Al + 3Ag₂S → 6Ag + Al₂S₃,

It is clear that 2 mol of Al react with 3 mol of Ag₂S to produce 1 mol of Ag and 1 mol of Al₂S₃.

Al reacts with Ag₂S with (2: 3) molar ratio.

So, 2.27 mol of Al reacts completely with 3.4 mol of Ag₂S with (2: 3) molar ratio.

The limiting reactant is Ag₂S.

The excess "left over" reactant is Al.

The reamining moles of excess reactant "Al" = 8.1 mol - 2.27 mol = 5.83 mol.

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3 years ago