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Scilla [17]
3 years ago
11

Suggest a way of improving

Chemistry
1 answer:
bezimeni [28]3 years ago
4 0

Explanation:

when water is boiling and when u cover it and increase the pressure it causes the increase in rate of condensation.

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monkey

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While in Europe, if you drive 125 km per day, how much money would you spend on gas in one week if gas costs 1.10 euros per lite
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A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe
nikitadnepr [17]

Answer:

\large \boxed{34.2\, ^{\circ}\text{C}}

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:  

For the metal:

m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}

For the water:

m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}

\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}

\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}

\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}

3 0
3 years ago
Which descriptions apply to emitted radiation? check all that apply?
Murrr4er [49]
<span>measurement in Ci/Bq

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What is the majority of 0.50 g of na dissolved in a 1.5 L solution
Oxana [17]

The correct answer is 0.014467 M.  

Molarity is defined as the number of moles present in a liter solution, that is, number of moles / liter solution.  

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Thus, 1 mole of Na contains 23.0 g

Now, x moles of Na contains 0.50 g

Moles = 0.50 × 1 / 23.0

Moles = 0.50 / 23.0  

= 0.0217 moles of Na

Molarity = Number of moles / liters of solution

= 0.0217 / 1.5

= 0.014467 M


8 0
4 years ago
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