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3 years ago

WILL GIVE BRAINLIEST IF SOMEONE ANSWERS THIS CORRECTLY ASAP!!!!

Chemistry

1 answer:

zhuklara [117]3 years ago

4 0

Answer:

it is burning of dry leaves in the garden

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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

A very flammable substance would not be used in children’s clothing? true or false

Alex73 [517]

True because certain fabric will catch on fire

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3 years ago

Question 21 of 25

RideAnS [48]

Answer:

Explanation:

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3 years ago

How many grams of NaCI per kilogram of water are present in a 2.7 molal aqueous solution

navik [9.2K]

Answer:

157.79 g

Explanation:

The definition of molality is:

molality = moles of solute / kilogram of solvent

This means that in a 2.7 molal solution, there are 2.7 moles of NaCl per kilogram of water.

So now w<u>e convert those 2.7 moles of NaCl to grams</u>, using its <em>molar mass</em>:

2.7 mol * 58.44 g/mol = 157.79 g

5 0

3 years ago

Charle's Law

Setler [38]

Convert temperature to Kelvin

225°C=498K
127°C=400K

Convert vol to L

400mL=0.4L

Apply Charles law

V1T_2=V2T_1
0.4(400)=498V_2
160=498V_2
V_2=0.32L=320mL

8 0

2 years ago

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*WILL GIVE BRAINLIEST IF SOMEONE ANSWERS THIS CORRECTLY ASAP!!!!*

WILL GIVE BRAINLIEST IF SOMEONE ANSWERS THIS CORRECTLY ASAP!!!!