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3 years ago

What is the mass in grams of a pure iron cube that has a volume of 4.60 cm3

1 answer:

5 0

In order to determine the mass of a substance given the volume, we require the density. The density of iron is 7.87 grams/cm³

Now,

Density = mass / volume

Mass = density * volume

Mass = 7.87 * 4.6

36.2 grams of iron are present in the cube

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The flamability (burns) is a physical change ?

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Flammability is a chemical change because when you burn something, it no longer has the same properties.

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3 years ago

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3 years ago

Starting with the stock solution of 6.0 M, how many milliliters of 6.0 M sulfuric acid are needed to make 450 mL of 1.2 M soluti

Lina20 [59]

<u>Answer:</u> The volume of stock solution needed is 90 mL

<u>Explanation:</u>

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the stock sulfuric acid solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted sulfuric acid solution

We are given:

$M_1=6.0M\\V_1=?mL\\M_2=1.2M\\V_2=450mL$

Putting values in above equation, we get:

$6.0\times V_1=1.2\times 450\\\\V_1=\frac{1.2\times 450}{6.0}=90mL$

Hence, the volume of stock solution needed is 90 mL

6 0

2 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

Jane has a weight of 500 N. The area of her two feet is 250 cm2. The pressure when she is standing is Pressure = 500/250 = 2 N/c

weeeeeb [17]

Explanation:

pressure=force/area

pressure=500/2

pressure=250N/cm²

8 0

3 years ago