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3 years ago

Which of the following substances would you expect to have a low melting point?

Chemistry

1 answer:

Lorico [155]3 years ago

5 0

The substance you would expect to have a low melting point would be carbon dioxide, so the correct answer would be A

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How to calculate magnitude of k

olasank [31]

Use Arrhenius equation:
k = A*exp(-Ea/RT)
We have:
1.35x10^2/s = A*exp(-85600/(8.314*298.15))
or: A = 1.342x10^17/s
It is a piece of cake to calculate:
k = 1.342x10^17*exp(-85600/(8.314*348.15))
= 1.92x10^4/s

6 0

3 years ago

A pupil adds 5 cm of 12 mol.dm sulphuric acids to make a 250 cmº solution. Calculate the

klemol [59]

Answer:

0.500 mol/dm³

Explanation:

Using the formula below;

CaVa = CbVb

Where;

Ca = concentration of acid (mol/dm³)

Cb = concentration of base (mol/dm³)

Va = volume of acid (cm³)

Vb = volume of base (cm³)

In accordance to the information provided in this question is;

Va = 5cm³

Vb = 250 cm³

Ca = 12 mol/dm³

Cb = ?

Using CaVa = CbVb

12 × 5 = Cb × 250

60 = 120Cb

Cb = 60/120

Cb = 0.500 mol/dm³

8 0

3 years ago

Which subshells (s, p, d, f, or g) can have electrons with the indicated magnetic quantum number (ml)?

amm1812

Answer:

=3 means is 3 or greater so that would be f and g subshells

=0 means is 0 or greater so that would be s, p, d, f and g subshells

=1 means is 1 or greater so that would be p, d, f, and g subshells

=4 means is 4 or greater so that would be g only

4 0

2 years ago

The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.

enot [183]

Answer:

$\large \boxed{\text{21.6 L}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 24.5

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}$

(b) Moles of CO₂

$\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}$

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37 °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

$\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}$

7 0

3 years ago

Calculate the molecular (formula) mass of each compound: (a) iron(ll) acetate tetrahydrate; (b) sulfur tetrachloride; (c) potass

Nutka1998 [239]

Answer:

a) Iron(ll) acetate tetrahydrate: 245,68 g/mol

b) Sulfur tetrachloride: 173,87 g/mol

c) Potassium ermanganate 158,034 g/mol

Explanation:

To solve this kind of exercises you must look for the number of atoms in each molecule first, then look on the periodic table the atom weight and the multiply the atom weight times the quantity of each atom. For instance:

The molecule of Iron(II) acetate itetrahydrate is (CH3COO)2Fe•4H2O, it means that you have:

2 atoms of carbon times the atomic weight of C (12.00g/mol)= 24g

14 atoms of Hidrogen times the atomic weight of H (1,00g/mol)= 14g

6 atoms of Oxigen times the atomic weight of O (16,0g/mol)= 96

2 atoms of Iron times the atomic weight of Fe (55,84g/mol)= 111,68g

At last, you only have to add the results: 24+14+96+11,68= 245,68g/mol. This example was for the first molecule.

See you,

5 0

3 years ago