What is the limiting reagent when 9.517 g of Fe is allowed to react with 19.34 g of water according to

The hydrogen-line emission spectrum includes a line at a wavelength of 434 nm. What is the energy of this radiation? (h= 6.626 x

Andrews [41]

Wavelength = 434nm = 434 x 10⁻⁹m
planck's constant = h= 6.626 x 10 ⁻³⁴ J
E =?
by using the formula;
E = hc /λ
value for c is 3 x 10⁸ m/s
E = (6.626 x 10 ⁻³⁴ J)(3 x 10⁸ m/s) / 434 x 10⁻⁹m
E = 1.9878 x 10⁻²⁵ / 434 x 10⁻⁹m
E = 4.58 x 10⁻¹⁹ joules

6 0

3 years ago

To a 25.00 mL volumetric flask, a lab technician adds a 0.150 g sample of a weak monoprotic acid, HA , and dilutes to the mark w

Elis [28]

Answer: The number of moles of weak acid is $4.24\times 10^{-3}$ moles.

Explanation:

To calculate the moles of KOH, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}\text{Volume of solution (in L)}}$

We are given:

Volume of solution = 43.81 mL = 0.04381 L (Conversion factor: 1L = 1000 mL)

Molarity of the solution = 0.0969 moles/ L

Putting values in above equation, we get:

$0.0969mol/L=\frac{\text{Moles of KOH}}{0.04381}\\\\\text{Moles of KOH}=4.24\times 10^{-3}mol$

The chemical reaction of weak monoprotic acid and KOH follows the equation:

$HA+KOH\rightarrow KA+H_2O$

By Stoichiometry of the reaction:

1 mole of KOH reacts with 1 mole of weak monoprotic acid.

So, $4.24\times 10^{-3}mol$ of KOH will react with = $\frac{1}{1}\times 4.24\times 10^{-3}=4.24\times 10^{-3}mol$ of weak monoprotic acid.

Hence, the number of moles of weak acid is $4.24\times 10^{-3}$ moles.

6 0

3 years ago

Which of the following uses a nonrenewable energy resource to generate energy? A. nuclear power plant B. wind farm C. fireplace

asambeis [7]

The answer is bimass power plant.

4 0

2 years ago

Glycerol has a molar mass of 92.09g/mol. Its percent composition is: 39.12% C, 8.75% H,

Sphinxa [80]

Answer:

Molecular formula => C₃H₈O₃

Explanation:

From the question given above, the following data were obtained:

Carbon (C) = 39.12%

Hydrogen (H) = 8.75%

Oxygen (O) = 51.12%

Molar mass of compound = 92.09 g/mol

Molecular formula =?

Next, we shall determine the empirical formula of the compound. This can be obtained as follow:

C = 39.12%

H = 8.75%

O = 51.12%

Divide by their molar mass

C = 39.12 / 12 = 3.26

H = 8.75 / 1 = 8.75

O = 51.12 / 16 = 3.195

Divide by the smallest

C = 3.26 / 3.195 = 1

H = 8.75 / 3.195 = 2.7

O = 3.195 / 3.195 = 1

Thus, the empirical formula is CH₂.₇O

Finally, we shall determine the molecular formula of the compound. This can be obtained as follow:

Empirical formula = CH₂.₇O

Molar mass of compound = 92.09 g/mol

Molecular formula =?

Molecular formula = Empirical formula × n

Molecular formula = [CH₂.₇O]ₙ

92.09 = [12 + (2.7×1) + 16] × n

92.09 = 30.7n

Divide both side by 30.7

n = 92.09 / 30.7

n = 3

Molecular formula = [CH₂.₇O]ₙ

Molecular formula = [CH₂.₇O]₃

Molecular formula = C₃H₈O₃

3 0

3 years ago

Write the balanced chemical equation between H2SO4H2SO4 and KOHKOH in aqueous solution. This is called a neutralization reaction

const2013 [10]

Answer:

0.185M sulfuric acid

Explanation:

Based on the reaction:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

1 mole of sulfuric acid reacts with 2 moles of KOH

Initial moles of H₂SO₄ and KOH are:

H₂SO₄: 0.750L ₓ (0.470mol / L) = 0.3525 moles of H₂SO₄

KOH: 0.700L ₓ (0.240mol / L) = 0.168 moles of KOH

The moles of sulfuric acis that react with KOH are:

0.168mol KOH ₓ (1 mole H₂SO₄ / 2 moles KOH) = 0.0840 moles of sulfuric acid.

Thus, moles that remain are:

0.3525moles - 0.0840 moles = 0.2685 moles of sulfuric acid remains

As total volume is 0.700L + 0.750L = 1.450L, concentration is:

0.2685mol / 1.450L = 0.185M sulfuric acid

8 0

3 years ago