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3 years ago

Diffrent Between Gas and Air

Chemistry

1 answer:

ollegr [7]3 years ago

3 0

Answer:

air is a group of different gases and gas is loose bound atoms of an element

Explanation:

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How many moles of hcl are present in 40.0 ml of a 0.035 m solution?

Elden [556K]

Answer:

0.0014 moles is present in 40cm³ of 0.035M of HCl solution

Explanation:

Molarity = 0.035M

V = 40.0mL

1mL = 1cm³

V = 40cm³

0.035 moles = 1000cm³

X moles is present in 40cm³

X = (40 * 0.035) / 1000

X = 0.0014moles

0.0014 moles is present in 40cm³ of solution

8 0

3 years ago

What are the two basic ways in which ions form from atoms?

andrew-mc [135]

Answer:

An ion is an atom or group of atoms that has an electric charge. When an atom loses an electron, it loses a negative charge and becomes a positive ion. When an atom gains an electron, it gains a negative charge and becomes a negative ion. a sodium ion forms when a sodium atom loses one electron and becomes positively charged. a chloride ion forms when a chlorine atom gains one electron and becomes negatively charged.

Hope this helps,

Arabella

4 0

3 years ago

What is the specific latent heat of fusion of ice if it takes 863 kJ to convert 4.6 kg of ice into water at 0 C?

charle [14.2K]

Answer:

The correct answer is 187.7 J/Jg.

Explanation:

The formula for finding the specific heat of fusion is,

Specific heat of fusion = Q/m

Here Q is the heat energy added, signified in kJ, and m is the mass of the object in kg.

Based on the given information, the heat energy added or Q is 869 kJ and the mass of the ice is 4.6 Kg

Now putting the values in the formula we get,

Specific heat of fusion = Q/m

Specific heat of fusion = 863 kJ / 4.6 Kg = 187.7 J/Kg

5 0

3 years ago

meriva

Answer:

Option B is correct. A nuclear alpha decay

Explanation:

Step 1

This equation is a nuclear reaction. So it can be an alpha decay or a beta decay

An α-particle is a helium nucleus. It contains 2 protons and 2 neutrons, for a mass number of 4.

During α-decay, an atomic nucleus emits an alpha particle. It transforms (or decays) into an atom with an atomic number 2 less and a mass number 4 less.

Thus, radium-226 decays through α-particle emission to form radon-222 according to the equation that is showed.

A Beta decay occurs when, in a nucleus with too many protons or too many neutrons, one of the protons or neutrons is transformed into the other.

Option B is correct. A nuclear alpha decay

6 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

Diffrent Between Gas and Air​

Diffrent Between Gas and Air