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2 years ago

How can you tell how many valence electrons an element has

Chemistry

1 answer:

Dimas [21]2 years ago

3 0

Answer:

the number of valence electrons is equal to the atom's main group number

Explanation:

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Describing Chemical Reactions

nadya68 [22]

Answer:

mass of Liquid 1 + mass of Liquid 2 = mass of solid + mass of gas

Explanation:

iquid 1 reacts with Liquid 2, producing a solid and a gas. Using this scenario, which supports the law of conservation

of mass?

EEEE

O mass of Liquid 1 + mass of solid = mass of Liquid 2 + mass of gas

O mass of Liquid 1 - mass of solid = mass of Liquid 2-mass of gas

O mass of Liquid 1 - mass of Liquid 2 = mass of solid + mass of gas

based on the law of the conservation of matter, the sum of the masses of the reactants must equal the sum of the masses of the products

so the correct answer is

mass of Liquid 1 + mass of Liquid 2 = mass of solid + mass of gas

7 0

3 years ago

You want to find the actual amount of reactants that are needed to get a desired amount of product. Which information is require

inysia [295]

Answer:

Explanation:

345tq325tqw26t42w6

4 0

4 years ago

If an atom has 7 protons, 8 neutrons and 9 electrons, what is the mass number?

Brums [2.3K]

Mass = protons + neutrons
7+8 = 15 g

3 0

3 years ago

Read 2 more answers

What can group numbers on the periodic table help you determine?

Ilia_Sergeevich [38]

Http://www.chem4kids.com/files/elem_pertable.html this should help.

4 0

3 years ago

How much energy (heat) is required to convert 248 g of water from 0oC to 154oC? Assume that the water begins as a liquid, that t

Arturiano [62]

Answer: The amount of heat required is 775.7 kJ

Explanation:

The processes involved in the given problem are:

$1.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\2.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\3.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\4.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(154^oC,427K)$

Now, we calculate the amount of heat released or absorbed in all the processes.

For process 1:

$q_1=m\times L_f$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 248 g

$L_f$ = latent heat of fusion = 334 J/g

Putting all the values in above equation, we get:

$q_1=248g\times 334J/g=84832J$

For process 2:

$q_2=m\times C_{p,l}\times (T_{2}-T_{1})$

where,

$q_2$ = amount of heat absorbed = ?

$C_{p,l}$ = specific heat of water = 4.184 J/g°C

m = mass of water = 248 g

$T_2$ = final temperature = $100^oC$

$T_1$ = initial temperature = $0^oC$

Putting all the values in above equation, we get:

$q_2=248g\times 4.184J/g^oC\times (100-0)^oC=103763.2J$

For process 3:

$q_3=m\times L_v$

where,

$q_3$ = amount of heat absorbed = ?

m = mass of water or ice = 248 g

$L_v$ = latent heat of vaporization = $40.79kJ/mol\times \frac{1000}{18}=2266.1J/g$ (Conversion factor used: 1 kJ = 1000 J and molar mass of water = 18 g/mol)