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3 years ago

When a substance accepts a proton (H+) what is it becoming? Is it becoming a negative or positive ?

Chemistry

1 answer:

d1i1m1o1n [39]3 years ago

6 0

Answer:

It’s becoming more positive, although it doesn’t inherently make it positive

Explanation:

Was the question worded like this? If so then the answer is def positive

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What is the molarity of a stock solution if 60 mL were used to make 150 ml<br>of a .5M solution?

olga2289 [7]

The molarity of the stock solution is 1.25 M.

<u>Explanation:</u>

We have to find the molarity of the stock solution using the law of volumetric analysis as,

V1M1 = V2M2

V1 = 150 ml

M1 = 0.5 M

V2 = 60 ml

M2 = ?

The above equation can be rearranged to get M2 as,

M2 = $$\frac{V1M1}{V2}$

Plugin the values as,

M2 = $$\frac{150 \times 0.5}{60}$

= 1.25 M

So the molarity of the stock solution is 1.25 M.

4 0

3 years ago

8. Select the lattice energy for rubidium chloride from the following data (in kJ/mol]

yKpoI14uk [10]

Answer:

Option C

Explanation:

The chemical reactions which are involved while solving this problem is there in the file attached and each chemical reaction is represented by a certain equation number

Lattice energy for rubidium chloride ( RbCl) is represented by the equation 6

Equation 1 represents the change in enthalpy for formation of RbCl

Equation 2 represents the sublimation reaction of rubidium

Equation 3 represents the ionization enthalpy of rubidium

Equation 4 represents the enthalpy of atomization of chlorine which means it describes the bond enthalpy of Cl2 molecule

Equation 5 represents the electron affinity of chlorine

To find the lattice energy for RbCl we have to use all the equations from 1 to 5 so that at last we get the equation 6

We have to perform operations such as

Equation 1 - equation 2 - equation 3 - equation 4 - equation 5

By performing these operations the intermediate compounds gets cancelled and at last we get equation 6

So Equation 1 ≡ ΔH $_{f}$ = -431 kJ/mol

Equation 2 ≡ Rb(s) ---> Rb(g) = 85.8 kJ/mol

Equation 3 ≡ IE1(Rb) = 397.5 kJ/mol

Equation 4 ≡ BE(Cl2) = 226 kJ/mol

Equation 5 ≡ Electron Affinity Cl = -332 kJ/mol

Value corresponding to the equation 6 will be the value of lattice energy of RbCl and the value is -695·3 kJ/mol

∴ Lattice energy for rubidium chloride is approximately -695 kJ/mol

4 0

3 years ago

Ne4ueva [31]

Answer:

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434 grams of F.

Explanation:

Volume of cylindrical reservoir = V

Radius of the cylindrical reservoir = r = d/2

d = diameter of the cylindrical reservoir = d = $4.50\times 10^1 m=45 m$

r = d/2 = 22.5 m

Depth of the reservoir = h = 10.0 m

$V=\pi r^2 h$

$=3.14\times (22.5 m)^2\times 10.0 m=15,896.25 m^3=15,896,250 L$

$1 m^3=1000 l$

Volume of water cylindrical reservoir : V

Density of water,d = 1 kg/L

Mass of water cylindrical reservoir = m

$m=d\times V=1 kg/L\times 15,896,250 L=15,896,250 kg$

1.6 kilogram of fluorine per million kilograms of water. (Given)

Concentration of fluorine in water = 1.6 kg/ 1000,000 kg of water

In 1000,000 kg of water = 1.6 kg of fluorine

Then $15,896,250 kg$ of water have x mass of fluorine:

$\frac{x}{15,896,250 kg\text{kg of water}}=\frac{1.6 kg}{1000,000 \text{kg of water}}$

$x=\frac{1.6 kg}{1000,000}\times 15,896,250 kg=25.434 kg$

15,896,250 kg water of contains mass 25.434 kg of fluorine.

25.434 kg = 25434 g

25,434 grams of fluorine should be added to give 1.6 ppm.

Percentage of fluorine in hydrogen hexafluorosilicate :

Molar mass hydrogen hexafluorosilicate = 144 g/mol

$F\%=\frac{6\times 19 g/mol}{144 g/mol}\times 100=79.16\%$

Total mass of hydrogen hexafluorosilicate = m'

$79.16\%=\frac{25,434 g}{m'}\times 100$

m' = 32,127.02 g

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434 grams of F.

5 0

3 years ago

A common anesthetic in dentistry known as nitronox consists of a mixture of dinitrogen oxide, n2o, and oxygen gas, o2, which is

Nesterboy [21]

Anesthetic in dentistry consists of a mixture of dinitrogen oxide (N₂O) and oxygen gas (O₂), which is administered through an inhaler over the nose. Total pressure of the mixture ( $P_{total}$ ) is sum of partial pressure of N₂O ( $p_{N_{2}O }$ ) and partial pressure of O₂, ( $p_{O_{2} }$ ).