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Klio2033 [76]
3 years ago
15

What is the mass of 1 mole of baking soda (sodium hydrogen carbonate) which has a formula of NaHCO ?​

Chemistry
1 answer:
aleksandr82 [10.1K]3 years ago
3 0

Answer:

<em>1 Mole = 84.007 g/mol</em>

Explanation:

Sodium bicarbonate (IUPAC name: sodium hydrogen carbonate), commonly known as baking soda or bicarbonate of soda, is a chemical compound with the formula NaHCO3

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Ionic and polar compounds usually dissolve in which solvents?
vampirchik [111]
Three of the intermolecular forces of attraction are roughly equal, the substances<span> will be </span>soluble in each other, <span>This means that </span>ionic<span> or </span>polar<span> solutes </span>dissolve<span> in </span><span>polar solvents</span>
3 0
3 years ago
Water is an amphoteric substance, meaning it can serve as both an acid and a base,
statuscvo [17]

Answer:

given statement is completely true

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5 0
3 years ago
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Electronic transitions (i.e., absorption of uv or visible light) of the conjugated molecule butadiene can be approximated using
dlinn [17]

Answer:

2.51 Angstroms

Explanation:

For a particle in a one dimensional box, the energy level, En, is given by the expression:

En = n²π² ħ² / 2ma²

where n is the energy level, ħ²  is Planck constant divided into 2π, m is the mass of the electron ( 9.1  x 10⁻³¹ Kg ), and a is the length of the one dimensional box.

We can calculate  the change in energy, ΔE, from n = 2 to n= 3  since we know the wavelength of the transition  ( ΔE = h c/λ ) and then substitute this value for the expresion of the ΔE for a particle in a box and solve for the  length a.

λ = 207 nm x 1 x 10⁻⁹ m/nm = 2.07 x 10⁻⁷ m      ( SI units )

ΔE = 6.626 x 10⁻³⁴ J·s x  3 x 10⁸ m/s  / 2.07 x 10⁻⁷ m

ΔE   = 9.60 x 10⁻¹⁹ J

ΔE(2⇒3) = ( 3 - 2 )  x  π² x ( 6.626 x 10⁻³⁴ J·s / 2π )² / ( 2 x 9.1 x 10⁻³¹ Kg x a² )

9.60 x 10⁻¹⁹ J  =  π² x( 6.626 x 10⁻³⁴ J·s / 2π )² / ( 2 x 9.1 x 10⁻³¹ Kg x a² )

⇒ a = 2.51 x 10⁻¹⁰ m

Converting to Angstroms:

a = 2.51 x 10⁻¹⁰ m x 1 x 10¹⁰ Angstrom / m = 2.51 Angstroms

6 0
4 years ago
50cm3 of sodium hydroxide solution was titrated against a solution of sulfuric acid. The concentration of the sodium hydroxide s
miskamm [114]

Answer:

49 g/L is the concentration of the acid

Explanation:

Firstly, we proceed to write the equation of reaction.

2NaOH + H2SO4 ——-> Na2SO4 + 2H2O

We can see that 1 mole of the base reacted with two moles of the acid.

kindly note that dm^3 is same as liter

Firstly, we need to get the concentration of the reacted sulphuric acid in g/L

we use the simple titration equation below;

CaVa/CbVb = Na/Nb

From the question;

Ca = ?

Va = 25 cm^3

Cb = 20 g/L

we convert this to concentration in mol/L

Mathematically, that is concentration in g/L divided by molar mass in g/mole

molar mass of NaOH = 40 g/mol

so we have; 20g/L / 40 = 0.5 mol/L

Vb = 50 cm^3

Na = 1

Nb = 2

Where C represents concentrations, V volumes and N , number of moles

Now, substitute the values;

Ca * 25/0.5 * 50 = 1/2

25Ca/25 = 0.5

So Ca = 0.5 mol/L

Now to get the concentration of H2SO4 in g/L

What we do is to multiply the concentration in mol/L by molar mass in g/mol

That would be 0.5 * 98 = 49 g/L

4 0
4 years ago
Plzzz help me out as I was not in class and I missed the lesson​
VARVARA [1.3K]

Answer:

Well why didn't you do it? Have a great day and I am so sorry for wasting your points I don't need them. I hope someone helps!! ;)

Explanation:

3 0
3 years ago
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