What is the main reason that the bounce height decrease.

How many grams of NaCl do you need to prepare 200 ml of a 5M solution? The molecular weight of NaCl is 58.4 g/mol. Smol 0.2L-1mo

Butoxors [25]

Answer:

The answer to your question is: 58.4 g of NaCl

Explanation:

Data

Volume = 200 ml = 0.2 l

Concentration = 5M

MW = 58.4 g

mass NaCl = ?

Formula

Molarity = (# of moles ) / volume

# of moles = Molarity x volume

# of moles = 5 x 0.2

# of moles = 1

58.4 g ---------------------- 1 mol

x --------------------- 1 mol

x = (1 x 58.4) / 1

x = 58.4 g of NaCl

4 0

3 years ago

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

Is a flashlight turned on kinetic or potential energy ?

kompoz [17]

Answer: potential energy

8 0

2 years ago

Which of the following is NOT true of white dwarfs?

liq [111]

Answer:

A white dwarf is what stars like the Sun become after they have exhausted their nuclear fuel. Near the end of its nuclear burning stage, this type of star expels most of its outer material, creating a planetary nebula. Only the hot core of the star remains. ... That means a white dwarf is 200,000 times as dense.

Explanation:

they are cold

they are about the size of Earth

these both are not true

6 0

3 years ago

Read 2 more answers

An industrial chemist is studying a sample of an unknown metal. Describe two ways he could change the metal physically and two w

Trava [24]

Answer:

Some of the physical changes used by the industrial chemist in order to identify it is by scratching it with other metals in order to find the hardness of it. Trying to deform it in order to find the malleability, and to heat it and measure the temperature in order to find the melting point.

Some of the chemical changes used by the industrial chemist in order to identify it is by inserting it in water to observe that whether it reacts with it or not, if the reaction is violent, then the metal belongs to either group I or group II. The other method is to insert it in acids of distinct strength and to observe its reaction. The metals belonging to the second group react briskly with acids. The other metals react gradually with acids and others are almost inert.

3 0

2 years ago