2 main reasons why we have seasons.

For the following reaction, 38.3 grams of sulfuric acid are allowed to react with 33.5 grams of calcium hydroxide sulfuric acid(

Likurg_2 [28]

Answer:

What is the maximum amount of calcium sulfate that can be formed? 53.1 grams CaSO4

What is the FORMULA for the limiting reagent? H2SO4

What amount of the excess reagent remains after the reaction is complete? 4.59 grams of Ca(OH)2

Explanation:

Step 1: Data given

Mass of sulfuric acid = 38.3 grams

Molar mass of H2SO4 = 98.08 g/mol

Mass of calcium hydroxide = 33.5 grams

Molar mass of Ca(OH)2 = 74.09 g/mol

Step 2: The balanced equation

H2SO4 + Ca(OH)2 → CaSO4 + 2H2O

Step 3: Calculate moles of H2SO4

moles H2SO4 = mass H2SO4 / molar mass H2SO4

moles H2SO4 = 38.3 grams / 98.08 g/mol

moles H2SO4 = 0.390 moles

Step 4: Calculate moles of Ca(OH)2

moles Ca(OH)2 = 33.5 grams / 74.09 g/mol

moles Ca(OH)2 =0.452 moles

Step 5: Calculate limiting reactant

For 1 mol H2SO4, we need 1 mol of Ca(OH)2 to produce, 1 mol of CaSO4 and 2 mol of H2O

H2SO4 is the limiting reactant. It will completely be consumed (0.390 moles).

Ca(OH)2 is in excess. There will be consumed 0.390 moles

There will remain 0.452 - 0.390 = 0.062 moles

This is 0.062 * 74.09 g/mol = 4.59 grams

Step 6: Calculate moles of calcium sulfate

For 1 mol H2SO4, we need 1 mol of Ca(OH)2 to produce, 1 mol of CaSO4 and 2 mol of H2O

For 0.390 moles of H2SO4, there will be produced 0.390 moles of CaSO4

Step 7: Calculate mass of CaSO4

Mass CaSO4 = moles CaSO4 * molar mass CaSO4

Mass CaSO4 = 0.390 moles * 136.14 g/mol

Mass of CaSO4 = 53.1 grams

7 0

3 years ago

Using the Bohr model, determine the energy, in joules, necessary to ionize a ground-state hydrogen atom. Show your calculations.

lord [1]

Answer:

The energy required to ionize the ground-state hydrogen atom is 2.18 x 10^-18 J or 13.6 eV.

Explanation:

To find the energy required to ionize ground-state hydrogen atom first we calculate the wavelength of photon required for this operation.

It is given by Bohr's Theory as:

1/λ = Rh (1/n1² - 1/n2²)

where,

λ = wavelength of photon

n1 = initial state = 1 (ground-state of hydrogen)

n2 = final state = ∞ (since, electron goes far away from atom after ionization)

Rh = Rhydberg's Constant = 1.097 x 10^7 /m

Therefore,

1/λ = (1.097 x 10^7 /m)(1/1² - 1/∞²)

λ = 9.115 x 10^-8 m = 91.15 nm

Now, for energy (E) we know that:

E = hc/λ

where,

h = Plank's Constant = 6.625 x 10^-34 J.s

c = speed of light = 3 x 10^8 m/s

Therefore,

E = (6.625 x 10^-34 J.s)(3 x 10^8 m/s)/(9.115 x 10^-8 m)

E = 2.18 x 10^-18 J

E = (2.18 x 10^-18 J)(1 eV/1.6 x 10^-19 J)

E = 13.6 eV

5 0

4 years ago

A cylindrical piece of metal is 4.5 dm in height with radius of 5.50 x 10^-5 km.

adell [148]

Answer:

a) $V=4.3x10^3mL$

b) $V=4.3x10^6mm^3$

c) $\rho=1.5x10^5g/L$

Explanation:

Hello,

a) In this case, the given height in cm is:

$h=4.5dm*\frac{1m}{10dm}* \frac{100cm}{1m}=45cm$

And the radius in cm is:

$r=5.50x10^{-5}km*\frac{1000m}{1km}*\frac{100cm}{1m}=5.5cm$

Thus, the volume in cubic centimeters which is also equal in mL (1cm³=mL) is:

$V=\pi (5.5cm)^2*45cm\\\\V=4.3x10^3cm^3=4.3x10^3mL$

b) In this case, the given height in mm is:

$h=4.5dm*\frac{1m}{10dm}* \frac{1000mm}{1m}=450mm$

And the radius in mm is:

$r=5.50x10^{-5}km*\frac{1000m}{1km}*\frac{1000mm}{1m}=55mm$

Thus, the volume in cubic millimeters is:

$V=\pi (55mm)^2*450mm\\\\V=4.3x10^6mm^3$

c) Finally, since 1000 mL equal 1 L, the required density in g/L turns out:

$\rho=\frac{m}{V}=\frac{6.54x10^5g}{4.3x10^3mL}*\frac{1000mL}{1L}\\ \\\rho=1.5x10^5g/L$

Best regards.

8 0

3 years ago

The standard molar heat of fusion of ice is 6020 j/mol. calculate q, w, and ∆e for melting 1.00 mol of ice at 0◦c and 1.00 atm p

zysi [14]

Answer : q = 6020 J, w = -6020 J, Δe = 0

Solution : Given,

Molar heat of fusion of ice = 6020 J/mole

Number of moles = 1 mole

Pressure = 1 atm

Molar heat of fusion : It is defined as the amount of energy required to melt 1 mole of a substance at its melting point. There is no temperature change.

The relation between heat and molar heat of fusion is,

$q=\Delta H_{fusion}(\frac{Mass}{\text{ Molar mass}})$ (in terms of mass)

or, $q=\Delta H_{fusion}\times Moles$ (in terms of moles)

Now we have to calculate the value of q.

$q=6020J/mole\times 1Mole=6020J$

When temperature is constant then the system behaves isothermally and Δe is a temperature dependent variable.

So, the value of $\Delta e=0$

Now we have to calculate the value of w.

Formula used : $\Delta e=q+w$

where, q is heat required, w is work done and $\Delta e$ is internal energy.

Now put all the given values in above formula, we get

$0=6020J+w$

w = -6020 J

Therefore, q = 6020 J, w = -6020 J, Δe = 0

3 0

3 years ago

Isotopes of the same element have different numbers of

Igoryamba

Isotopes of the same element have different numbers of
A protons
B neutrons
C neutrons and electrons
D protons and electrons
the answer is B neutrons

3 0

3 years ago