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Serhud [2]
3 years ago
5

Please help with sodium and fluorine

Chemistry
1 answer:
Alexxandr [17]3 years ago
8 0

Answer:

Sorry I wish I could help u.. I think this picture is correct

Explanation:

You might be interested in
There are three isotopes of X element (X):
Nutka1998 [239]

Answer:

Average atomic mass = 17.5 amu.

Explanation:

Given data:

X-17 isotope = atomic mass17.2 amu, abundance:78.99%

X-18isotope =  atomic mass 18.1 amu, abundance 10.00%

X-19isotope = atomic mass:19.1 amu, abundance: 11.01%

Average atomic mass of X = ?

Solution:

Average atomic mass  = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass)  / 100

Average atomic mass  = (78.99×17.2)+(10.00×18.1) +(11.01+ 19.1) /100

Average atomic mass =  1358.628 + 181 +210.291 / 100

Average atomic mass  = 1749.919 / 100

Average atomic mass = 17.5 amu.

8 0
3 years ago
Why do most atoms form chemical bonds?
zloy xaker [14]

Answer:

Why do most atoms form chemical bonds? They want a full outer shell of electrons, so the lose, gain, or share electrons with other elements, forming compounds, until they have 8 valence electrons and become stable. Double and triple covalent bonds that have greater bond energy and are shorter than single bonds.

Explanation: HOPE THIS HELPS YOU..

3 0
3 years ago
The size (radius) of an oxygen molecule is about 2.0 ×10−10m. Make a rough estimate of the pressure at which the finite volume o
belka [17]

Answer:

Explanation:

We can calculate the volume  of the oxygen molecule as the radius of oxygen molecule is given as 2×10⁻¹⁰m.

We know that volume=4/3×πr³

volume =4/3×π(2.0×10⁻¹⁰m)³

volume=33.40×10⁻³⁰m³

Volume of oxygen molecule=33.40×10⁻³⁰m³

we know the ideal gas equation as:

PV=nRT

k=R/Na

R=k×Na

PV=n×k×Na×T

n×Na=N

PV=Nkt

p is pressure of gas

v is volume  of gas

T is temperature of gas

N is numbetr of molecules

Na is avagadros number

k is boltzmann constant =1.38×10⁻²³J/K

R is real gas constant

So to calculate pressure using the  formula;

PV=NkT

P=NkT/V

Since there is only one molecule of oxygen so N=1

P=[1×1.38×10⁻²³J/K×300]/[33.40×10⁻³⁰m³

p=12.39×10⁷Pascal

8 0
3 years ago
Suppose you need of Grade 70 tow chain, which has a diameter of and weighs , to tow a car. How would you calculate the mass of t
BabaBlast [244]

Answer: check explanation

Explanation:

In this question we are to find mass. In order to calculate the Mass, We need the values of two parameters, that is, the values given for the grade tow chain, and the value given for the mass per length.

Assuming the mass per length is 3 Kilogram per metre(kg/m) and the grade 70 tow chain length is 5 metre(m).

Therefore, the formula for calculating mass of the chain is given below;

Mass of the chain= mass per unit length(kg/m) × length ---------------------------------------------------------------------------------------------------------------------(1).

Mass of the chain= 3 kg/m × 5 m.

Mass of the chain= 15 kg.

7 0
4 years ago
CaBr + KOH – Ca(OH), + KBr (balance first) What mass, in grams, of
neonofarm [45]

Answer:

129.73 g of CaBr₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

CaBr₂ + 2KOH –> Ca(OH)₂ + 2KBr

Next, we shall determine the mass of CaBr₂ that reacted and the mass of Ca(OH)₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of CaBr₂ = 40 + (80×2)

= 40 + 160

= 200 g/mol

Mass of CaBr₂ from the balanced equation = 1 × 200 = 200 g

Molar mass of Ca(OH)₂ = 40 + 2(16 + 1)

= 40 + 2(17)

= 40 + 34

= 74 g/mol

Mass of Ca(OH)₂ from the balanced equation = 1 × 74 = 74 g

SUMMARY :

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Finally, we shall determine the mass of CaBr₂ that react when 48 g of Ca(OH)₂ were produced. This can be obtained as follow:

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Therefore, Xg of CaBr₂ will react to produce 48 g of Ca(OH)₂ i.e

Xg of CaBr₂ = (200 × 48)/74

Xg of CaBr₂ = 129.73 g

Thus, 129.73 g of CaBr₂ were consumed.

6 0
3 years ago
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