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2 years ago

The total number of valence electrons in a molecule of sof2 is ________.

1 answer:

7 0

Answer:
There are 26 valence electrons in SOF₂.

Explanation:
Valence electrons are calculated by adding the valence electrons present in valence shell of each element present in SOF₂. So,

# of Valence e⁻s in S = 6

# of Valence e⁻s in O = 6

# of Valence e⁻s in F = 7

# of Valence e⁻s in F = 7

Now, add all valence electrons,

= 6 + 6 + 7 + 7

= 26 Valence Electrons

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I: Current
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3 years ago

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You water three sunflower plants with salt water. Each plant receives a different concentration of salt solutions. A fourth plan

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Answer:

Please please please mark me brainlest!

Explanation:

IV: salt solution

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3 years ago

You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

Now put all the given values in the above formula, we get:

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

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2 years ago

How many molecules of CBr4 are in 250 grams of CBr4

Kazeer [188]

Answer:- $4.54*10^2^3$ molecules.

Solution:- The grams of tetrabromomethane are given and it asks to calculate the number of molecules.

It is a two step unit conversion problem. In the first step, grams are converted to moles on dividing the grams by molar mass.

In second step, the moles are converted to molecules on multiplying by Avogadro number.

Molar mass of $CBr_4$ = 12+4(79.9) = 331.6 g per mol

let's make the set up using dimensional analysis:

$250g(\frac{1mol}{331.6g})(\frac{6.022*10^2^3molecules}{1mol})$

= $4.54*10^2^3$ molecules

So, there will be $4.54*10^2^3$ molecules in 250 grams of $CBr_4$ .

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