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3 years ago

5. What plate boundary could form a mountain chain of sedimentary rock?

Chemistry

1 answer:

Ivanshal [37]3 years ago

3 0

b) Convergent boundary of 2 continental crust

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The radius of an atom of gold (Au) is about 1.35 Å.

Eduardwww [97]

Two radius of an atom is equal to the diameter. Adding up all the diameter of the atoms, it should be equal to 9.5 mm. Therefore, we simply convert the units to the same units then divide 1.35 A to 9.5 mm. We calculate as follows:

no. of atoms = 0.0095 m / 1.35x10^-9 m = 7037037 atoms

Hope this answers the question. Have a nice day.

6 0

3 years ago

Read 2 more answers

Which of the following elements would make a 2- ion? A. N B. O C. Si D. Mg

nataly862011 [7]

Answer:

oxygen

Explanation:

because the 2nd shell is not complete which is suppose to be 8 and since oxygen is 8 it first shell is 2 which is complete and the second shell which is 6 is not complete because we all know that 2+6=8 but the standard shell is

K-2

L-8

M-8

3 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

A gas fills a balloon at a temperature of 27 C and 101.325 kPa of pressure. What will the pressure of the balloon be if the gas

JulijaS [17]

135.1‬kPa

Explanation:

Given parameters:

T1 = 27°C

P1 = 101.325 kPa

T2 = 127°C

Unknown:

P2 = ?

Solution:

Using a derivative of the combined gas law where we assume that the gas has a constant volume, we can solve for the unknown.

At constant volume:

$\frac{P1}{T1} = \frac{P2}{T2}$

P1 is the initial pressure

T1 is the initial temperature

P2 is the final pressure

T2 is the final temperature

Take the given temperature to K

T1 = 27 + 273 = 300K

T2 = 127 + 273 = 400K

Input the variables:

$\frac{101.325}{300} = \frac{P2}{400}$

P2 = 135.1‬kPa

learn more:

Boyle's law brainly.com/question/8928288

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3 0

3 years ago

Compare the type of change that occurs when an iron bar rusts and when a substance freezes.

Ivanshal [37]

Explanation:

when an iron bar rust is an example of a chemical change in which a new substance is formed and the change is not easily reversible.for iron to rust moisture and air must be present.while when a substance freezes,it can be easily reversed through melting and no new substance is formed.this change is termed a physical change.

5 0

2 years ago