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SVETLANKA909090 [29]
3 years ago
7

How do I find x mm? This is for my chemistry class.

Chemistry
1 answer:
Evgesh-ka [11]3 years ago
7 0
Im guessing it might be 98.4x0.58, when you rearrange the pressure formula.
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What is the name of a loop that electricity flows through?
aleksley [76]

Answer:

an electric circuit so 1

Explanation:

5 0
3 years ago
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If you have a gold brick that is 2cm by 3cm by 4cm and has density of 19.3g/cm3, what is its mass?
natulia [17]
Volume = a x a x a  

V = 2 cm x 3 cm x 4 cm => 24 cm³

Density = 19.3 g/cm³

Mass = ?

Therefore:

m = D x V

m = 19.3 x 24

m = 463.2 g
4 0
3 years ago
For each element, predict where the "jump " occurs for successive ionization energies. (For example, does the jump occur between
vichka [17]

Answer:

A jump occurs when a core electron is removed.

Explanation:

A jump in ionization energy occurs when a core electron is removed. A large jump in the ionization energy easily be seen from the electronic configuration of an element.

For Beryllium, the electronic configuration of is 1s2 2s2.

There are two valence electrons in the outermost shell hence the ionization energy data for beryllium will show a sudden jump or increase in going from the second to the third ionization energy owing to the removal of a core electron

The electronic configuration for Nitrogen is 1s2 2s2 2p3. Five valence electrons are found in the outermost shell so the ionization energy data for nitrogen will show a sudden jump or increase in going from the fifth to sixth ionization energy because of the removal of a core electron

The electronic configuration of oxygen is 1s2 2s2 2p4. There are six valence electrons hence ionization energy for oxygen atom will show a sudden jump or increase in going from the sixth to the seventh ionization energy because of the removal of a core electron

The electronic configuration of Lithium is 1s2 2s1

There is one valence electron in its outermost shell so its ionization energy data will show a sudden jump or increase in going from the first to the second ionization energy because of the removal of a core electron.

8 0
4 years ago
Ethane, C2H6, can be formed by reacting acetylene, C2H2, with hydrogen gas as follows: C2H2(g) H2(g) ⇌ C2H6(g) Exothermic What c
Pavel [41]

Answer:

Equilibrium shifts to the right

Explanation:

An exothermic reaction is one in which temperature is released to the environment. Hence, if the reaction vessel housing an exothermic reaction is touched after reaction completion, we will notice that the reaction vessel e.g beaker is hot.

To consider the equilibrium response to temperature changes, we need to consider if the reaction is exothermic or endothermic. In the case of this particular question, it has been established that the reaction is exothermic.

Heat is released to the surroundings as the reactants are at a higher energy level compared to the products. Hence, increasing the temperature will favor the formation of more reactants and as such, the equilibrium position will shift to the left to pave way for the formation of more reactants. Thus , more acetylene and hydrogen would be yielded

8 0
3 years ago
Determine the molality of a solution of methanol dissolved in ethanol for which the mole fraction of methanol is 0.135. Give you
Alja [10]

<u>Answer:</u> The molality of the solution is 0.11 m

<u>Explanation:</u>

We are given:

Mole fraction of methanol = 0.135

This means that 0.135 moles of methanol is present in 1 mole of a solution

Moles of ethanol = 1 - 0.135 = 0.865 moles

To calculate the mass for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of ethanol = 0.865 moles

Molar mass of ethanol = 46 g/mol

0.865mol=\frac{\text{Mass of ethanol}}{46g/mol}\\\\\text{Mass of ethanol}=(0.865mol\times 46g/mol}=39.79g

To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

Where,

m_{solute} = Given mass of solute (methanol) = 0.135 g

M_{solute} = Molar mass of solute (methanol) = 32 g/mol

W_{solvent} = Mass of solvent (ethanol) = 39.79 g

Putting values in above equation, we get:

\text{Molality of methanol}=\frac{0.135\times 1000}{32\times 39.79}\\\\\text{Molality of methanol}=0.106m\approx 0.11m

Hence, the molality of the solution is 0.11 m

6 0
3 years ago
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