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3 years ago

Which of the following is a characteristic of a good roasting pan?

Chemistry

1 answer:

lbvjy [14]3 years ago

4 0

Answer:

i can't understand the question

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There are two different compounds of sulfur and fluorine.

weqwewe [10]

Atomic mass of F: 19.0 g/mol

Atomic mass of S: 32.1 g/mol

1.18 g F = [1.18 g / 19.0 g / mol] = 0.062 mol F

1 g S = 1 g/ 32.1 g/mol = 0.031 mol S

Divide by 0.031

0.062 mol F / 0.031 = 2 mol F

0.031 mol S / 0.031 = 1 mol S

SF2 Then X = 2

Verification:
F2 = 2*19.0 g = 38 g F
S = 32.1 g

36 gF / 32.1 g S = 1.18 g F / g S

7 0

3 years ago

What word describes the splitting of a larger atom into smaller ones?

Westkost [7]

Fission is a term that means when a large atom breaks into smaller atoms. It’s counterpart, Fusion, is the opposite, where small atoms fuse into larger atoms.

4 0

3 years ago

Read 2 more answers

In a fuel cell there is direct conversion of mechanical energy into electricity.

rewona [7]

Answer:

Explanation:

A fuel cell uses both hydrogen and oxygen as fuels. At the cathode oxygen is reduced to water according to the equation

O2+4H^+ +4e ------>2H2O

And hydrogen is oxidized at the anode:

2H2------>4H^+ +4e

Fuel cells produce only water as a by product hence they are environment friendly

5 0

3 years ago

Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec

Bess [88]

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0$

Let energy change be E for 1 atom.

$E=E_{\infty}-E_1=0-(-13.6 eV)=13.6 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol$

$E'=1,312.17 kJ/mol$

The energy required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ , $B^{4+}$ atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-340eV)=340 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 340eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol$

$E'=32,804.31 kJ/mol$

The energy required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ , $Li^{2+}$ atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol$

$E'=11,809.55 kJ/mol$

The energy required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ , $Mn^{24+}$ atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol$

$E'=820,107.88 kJ/mol$

The energy required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0

4 years ago

Balance the following equation: H20 --> H2 + O2

MakcuM [25]

2H2O = 2H2 + O2.

<h3><u>Explanation</u>:</h3>

Balancing equations is very essential because of the fact that it represents the stoichiometric quantities of the reactants needed to react to form the product. The ratio of the weights of reactant and product are also very well understood from this.

Here in this equation, the water is broken into hydrogen and oxygen. The balanced reaction is

2H2O = 2H2 + O2.

Two moles of water is broken down into 2 moles of hydrogen and one mole of oxygen.

5 0

3 years ago

Which of the following is a characteristic of a good roasting pan? ​

Which of the following is a characteristic of a good roasting pan?