Where does the ocean get the most intense heat from the sun .

An alkaline battery produces electrical energy according to the following equation.

Gnesinka [82]

Answer:

Part a: limiting reactant MnO₂

Part b: 12.43 g of Zn(OH)₂

Explanation:

Part a

Data given:

mass of Zn = 37.0 g

mass of MnO₂ = 21.5 g

Limiting reactant = ?

Given Reaction

Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

To determine the limiting reactant first we will look at the reaction

Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

1 mol 2 mol

Convert moles to mass

molar mass of Zn = 65.4 g/mol

Mass of MnO₂ = 55 + 2 (16) = 55 + 32 = 87 g/mol

So,

Zn(s) + 2 MnO₂(s) + H₂O(l) ----------> Zn(OH)₂(s) + Mn₂O₃(s)

1 mol (65.4 g/mol) 2 mol (87 g/mol)

65 g 174 g

So its clear from the reaction that 65 g Zn react with 174 g of MnO₂.

now if we look at the given amounts the amount MnO₂ is less then the amount of Zn but in actual calculation amount of MnO₂ is more then amount of zinc.

So, for MnO₂ if we calculate the needed amount of zinc

So apply unity formula

65 g Zn react ≅ 174 g of MnO₂

X g of Zn ≅ 21.5 g of MnO₂

Do cross multiplication

X g Zn react = 65 g x 21.5 g / 174 g

X g of Zn ≅ 8.032 g

So, 8.032 g of zinc will react out of 37.0 grams. the remaining will be in excess.

So MnO₂ will be consumed completely an it will be limiting reactant.

____________

part b

Data given:

mass of Zn = 37.0 g

mass of MnO₂ = 21.5 g

Mass of Zn(OH)₂ produced = ?

Given Reaction

Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

Solution:

As from the part A we come to know that MnO₂ is limiting reactant, so the amount of Zn(OH)₂ will depend on the amount of MnO₂.

So first we convert mass of MnO₂ to moles

Formula Used

no. of moles = mass in grams / molar mass

Mass of MnO₂ = 55 + 2 (16)

Mass of MnO₂ = 55 + 32 = 87 g/mol

Put values in the above equation

no. of moles = 21.5 g / 87 g/mol

no. of moles = 0.25 moles

Now,

Look at the reaction

Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

2 mol 1 mol

So its clear from the reaction that 2 mole of MnO₂ gives 1 mole of Zn(OH)₂

then how many moles of Zn(OH)₂ will be produce by 0.25 moles of MnO₂

So.

apply unity formula

2 mol of MnO₂ ≅ 1 mole of Zn(OH)₂

0.25 moles of MnO₂ ≅ X mole of Zn(OH)₂

Do cross multiplication

X mole of Zn(OH)₂ = 1 mole x 0.25 mol x / 2 mol

X mole of Zn(OH)₂ ≅ 0.125

Now Conver moles of Zn(OH)₂ to mass

Formula used

mass in grams = no. of moles x molar mass

Molar mass of Zn(OH)₂

Molar mass of Zn(OH)₂ = 65.4 + 2 (16 + 1)

Molar mass of Zn(OH)₂ = 65.4 + 2 (17)

Molar mass of Zn(OH)₂ = 65.4 + 34 = 99.4 g/mol

Put values in above equation

mass in grams = 0.125 mol x 99.4 g/mol

mass in grams = 12.43 g

So,

12.43 g of Zn(OH)₂ will be produce.

5 0

2 years ago

A 3.00-L flask is filled with gaseous ammonia, NH3. The gas pressure measured at 27.0 ∘C is 2.55 atm . Assuming ideal gas behavi

Whitepunk [10]

Answer : The mass of ammonia present in the flask in three significant figures are, 5.28 grams.

Solution :

Using ideal gas equation,

$PV=nRT\\\\PV=\frac{w}{M}\times RT$

where,

n = number of moles of gas

w = mass of ammonia gas = ?

P = pressure of the ammonia gas = 2.55 atm

T = temperature of the ammonia gas = $27^oC=273+27=300K$

M = molar mass of ammonia gas = 17 g/mole

R = gas constant = 0.0821 L.atm/mole.K

V = volume of ammonia gas = 3.00 L

Now put all the given values in the above equation, we get the mass of ammonia gas.

$(2.55atm)\times (3.00L)=\frac{w}{17g/mole}\times (0.0821L.atm/mole.K)\times (300K)$

$w=5.28g$

Therefore, the mass of ammonia present in the flask in three significant figures are, 5.28 grams.

8 0

3 years ago

Answer each of the following questions with increases, decreases, or does not change.

Nitella [24]

Answer:

a) increases

b) decreases

c) does not change

d) increases

Explanation:

The vapour pressure of a liquid is dependent on;

I) the magnitude of intermolecular forces

II) the temperature of the liquid

Hence, when any of these increases, the vapour pressure increases likewise.

Similarly, the boiling point of a liquid depends on the magnitude of intermolecular forces present because as intermolecular forces increases, more energy is required to break intermolecular bonds.

Lastly, increase in surface area of a liquid does not really affect it's vapour pressure.

7 0

2 years ago

What is the threshold frequency ν0 of cesium?

Montano1993 [528]

I think this is what you're after:

Cs(g) → Cs^+ + e⁻ ΔHIP = 375.7 kJ mol^-1 [1]

Convert to J and divide by the Avogadro Const to give E in J per photon

E = 375700/6.022×10^23 = 6.239×10^-19 J

Plank relationship E = h×ν E in J ν = frequency (Hz s-1)

Planck constant h = 6.626×10^-34 J s

6.239×10^-19 = (6.626×10^-34)ν

ν = 9.42×10^14 s^-1 (Hz)

IP are usually given in ev Cs 3.894 eV

E = 3.894×1.60×10^-19 = 6.230×10^-19 J per photon

4 0

2 years ago

Naturally occurring element X exists in three isotopic forms: X-28 (27.979 amu, 92.21% abundance), X-29 (28.976 amu 4.70% abunda

bezimeni [28]

Answer: The average atomic mass of X is 28.09 amu

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

For isotope 1:

Mass of isotope 1 = 27.979 amu

Percentage abundance of isotope 1 = 92.21 %

Fractional abundance of isotope 1 = 0.9212

For isotope 2:

Mass of isotope 2 = 28.976 amu

Percentage abundance of isotope 2 = 4.70 %

Fractional abundance of isotope 2 = 0.0470

For isotope 3:

Mass of isotope 3 = 29.974 amu

Percentage abundance of isotope 3 = 3.09 %

Fractional abundance of isotope 3 = 0.0309

Putting values in equation 1, we get:

$\text{Average atomic mass of X}=[(27.979\times 0.9212)+(28.976\times 0.0470)+(29.974\times 0.0309)]$

$\text{Average atomic mass of X}=28.09amu$

Hence, the average atomic mass of X is 28.09 amu

4 0

3 years ago