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xenn [34]
2 years ago
15

Na3X, Enter the group number of X?

Chemistry
2 answers:
ICE Princess25 [194]2 years ago
8 0

Answer : The group number of X is 15.

Explanation :

In Na_3X, 'Na' is a sodium metal and 'X' is an unknown element.

First we have to determine the oxidation state of element 'X'.

Let the oxidation state of element 'X' be, 'a'.

As we know that sodium is an alkali metal which belongs to group 1 and has (+1) charge.

3(+1)+a=0\\\\+3+a=0\\\\a=-3

That means, the oxidation state or charge on 'X' is (-3) and the group 15 elements (nitrogen, phosphorous, arsenic, antimony and bismuth) have five electrons in valence shell and thus require three more electrons to acquire the nearest noble gas configuration. Thus, the group 15 elements have the tendency to show (-3) oxidation state.

Hence, the group number of X is 15.

Katen [24]2 years ago
4 0
X will be in group 5, since if you exchange the valencies of Na with any element on group 5, you will get Na3X
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Answer:

Drug calculation

If we have 45g of clobetasol  = 0.05%w/w

Then what mass in g of clobetasol is in 0.03%w/w = 45 x 0.03/0.05 =27g

It means that 27g of clobetasol must be added to change the drug strength to 0.03% w/w

3 0
2 years ago
What is the overall balanced equation for the precipitation reaction occurring between silver nitrate and calcium bromide? hints
aleksklad [387]
The answer:

<span>the overall balanced equation for the precipitation reaction occurring between silver nitrate and calcium bromide:
</span>CaBr2 + AgNO3-----------> AgBr2 + CaNO3, so among the given choices, the
true answe is 
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4 0
2 years ago
g When 2.50 g of methane (CH4) burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion (in kJ) per mole
Anna [14]

Answer:

-800 kJ/mol

Explanation:

To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).

First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:

Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol

moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄

Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:

ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol

Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).

3 0
2 years ago
The Lewis structure for CO2 has a central The Lewis structure for C O 2 has a central blank atom attached to blank atoms.
kirza4 [7]

Answer:

See the explanation

Explanation:

1) The Lewis structure for  CO_2 has a central Carbon<em> </em>atom attached to Oxygen atoms.

In the CO_2  we will have a structure:  O=C=O the <u>central atom</u> "carbon" we will have <u>2 sigma bonds and 2 pi bonds</u>, therefore, we have an <u>Sp hybridization</u>. For O we have <u>1 pi and 1 sigma bond</u>, therefore, we have an <u>Sp2 hybridization</u>.

2) These atoms are held together by <u>double bonds.</u>

<u></u>

Again in the structure of CO_2: O=C=O we only have double bonds.

3. Carbon dioxide has a Carbon dioxide has a <u>Linear</u> electron geometry.

Due to the double bonds we have to have a linear structure because in this geometry the atoms will be further apart from each other.

4. The carbon atom is <u>Sp</u> hybridized.

We will have for carbon 2 pi bonds, so we will have an <u>Sp</u> hybridization.

5. Carbon dioxide has two Carbon dioxide has two C(p) - O(p) π bonds and two C(sp) - O(Sp2) σ bonds.

(See figures)

Figure 1: Carbon hybridization

Figure 2: Oxygen hybridization

6 0
2 years ago
The percent composition by mass of a compound is 76.0% c, 12.8% h, and 11.2% o. the molar mass of this compound is 284.5 g/mol.
Leokris [45]
You have a few steps to solve this one. First, we'll find the molar mass by percentage of each element in the molecule. Then, we'll divide each of those relative masses by the atomic mass of each element. The number of times the mass divides into the relative mass is the number of atoms of that element in the molecule:

C: 284.5 x .76 = 216.22
H: 284.5 x .128= 36.416
O: 284.5 x .112 = 31.864.

Now we divide out each element's atomic mass (from the periodic table). it's okay if they're approximated from the decimal answer.
C: 216.22 ÷ 12.011 ≈ 18
H: 36.416 ÷ 1.008 ≈36
O: 31.864 ÷ 15.999 ≈ 2

Therefore, the molecular formula is C18H36O2. 

The empirical formula would be found by dividing out all factors of those subscript numbers. In our case, all of them can be divided by 2. The empirical formula would be C9H18O




7 0
2 years ago
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