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3 years ago

Define atomic number and atomic mass

Chemistry

1 answer:

Akimi4 [234]3 years ago

3 0

Answer:

Atomic number ---> number of protons in the nucleus of an atom

Atomic mass ----> the total mass of protons, neutrons, and electrons in an atom

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Answer the four questions to figure out the four digit code

denis-greek [22]

Answer:

question 1 =c

question 2 =a

question 3 =d

question 4 =b

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2 years ago

Isopropyl methyl ether is slightly soluble with water because the oxygen atom of ethers with three or fewer carbon atoms can for

kondaur [170]

Answer:

True

Hydrogen bond is a partial intermolecular bonding interaction between a lone pair on an electron rich donor atom, particularly the second-row elements nitrogen (N), oxygen (O), or fluorine (F), and the antibonding orbital of a bond between hydrogen (H) and a more

electronegative atom or group. Such an interacting system is generally denoted Dn–H···Ac, where the solid line denotes a polar covalent bond, and the dotted or dashed line indicates the hydrogen bond. The use of three centered dots for the hydrogen bond is specifically recommended by the IUPAC. While hydrogen bonding has both covalence and electrostatic contributions, and the degrees to which they contribute are currently debated, the present evidence strongly implies that the primary contribution is covelant.

Hydrogen bonds can be intermolecular (occurring between separate molecules) or

intramolecular (occurring among parts of the same molecule)

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4 years ago

Pentaborane−9 (B5H9) is a colorless, highly reactive liquid that will burst into flames when exposed to oxygen. The reaction is

mote1985 [20]

Answer : The heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The equilibrium reaction follows:

$2B_5H_9(l)+12O_2(g)\rightleftharpoons 5B_2O_3(s)+9H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(n_{(B_2O_3)}\times \Delta H^o_f_{(B_2O_3)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(B_5H_9)}\times \Delta H^o_f_{(B_5H_9)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]$

We are given:

$\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(5\times -1271.94)+(9\times -285.83)]-[(2\times 73.2)+(12\times 0)]=-9078.57kJ/mol$

Now we have to calculate the heat released per gram of the compound reacted with oxygen.

From the reaction we conclude that,

As, 2 moles of compound released heat = -9078.57 kJ

So, 1 moles of compound released heat = $\frac{-9078.57}{2}=-4539.28kJ$

For per gram of compound:

Molar mass of $B_5H_9$ = 63.12 g/mole

$\Delta H^o_{rxn}=\frac{-4539.28}{63.12}=-71.915kJ/g$

Therefore, the heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

7 0

3 years ago

What is the relationship between a front and cloud formation?

denis-greek [22]

Answer:

Cold air is more dense than warm air, so when a warm air mass meets a cold air mass, the cold air ends up below the warm air. Once the air has risen, it cools and clouds can form. Weather fronts can cause clouds to form.

5 0

3 years ago

Read 2 more answers

Help please ASAP it’s 100 points I’ll mark you as brainlister but give a lil explation so I know you not cappin

alexdok [17]

Answer:

the answer to this question is option B mark me as brainliest

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3 years ago