What is the molar mass of PbSO4

A hydrate of CoCl2 has a mass of 17.40 g before heating. After heating, the mass of the anhydrous compound is found to be 10.27

melisa1 [442]

Mass CoCl2 = 10.27 g
moles CoCl2 = 10.27 g/ 129.839 g/mol=0.07910

mass water = 17.40 - 10.27=7.13 g
moles water = 7.13 / 18.02 g/mol=0.396

0.396/ 0.07910=5

CoCl2 * 5 H2O

moles CaF2 = 85.8 g/ 78.0748 g/mol=1.10
moles Ca = 1.10
mass Ca = 1.10 x 40.078 g/mol=44.1 g
V = 44.1 / 1.55 =28.5 mL

Greetings!!

5 0

3 years ago

Carbon dioxide is released into the atmosphere by _____.

kirill115 [55]

Answer:

4 a minor but very important component of the atmosphere corban dioxide is released through natural processes such as reputation.;

Explanation:

a minor but very important component of the atmosphere corban dioxide is released through natural processes such as reputation

3 0

2 years ago

Using the systematic approach for equilibrium problems, calculate the pH of 0.05 M HOCl. Ka= 3.0*10-8 Group of answer choices 3.

SVEN [57.7K]

Answer:

The pH is equal to 4.41

Explanation:

Since HClO is a weak acid, its dissociation in aqueous medium is:

HClO ⇄ ClO- + H+

start: 0.05 0 0

change -x +x +x

balance 0.05-x x x

As it is a weak acid it dissociates very little, in its ClO- and H + ions, so the change is negative, where x is a degree of dissociation.

the acidity constant when equilibrium is reached is equal to:

$Ka=\frac{[ClO-]*[H+]}{[HClO]}=\frac{x*x}{0.05-x}=3x10^{-8}$

The 0.05-x fraction can be approximated to 0.05, because the ionized fraction (x) is very small, therefore we have:

$3x10^{-8}=\frac{x^{2} }{0.05}$

clearing the x and calculating its value we have:

$x=3.87x10^{-5}=[H+]=[ClO-]$

the pH can be calculated by:

$pH=-log[H+]=-log[3.87x10^{-5}]=4.41$

7 0

3 years ago

The activation energy of an uncatalyzed reaction is 95kJ/mol. The addition of a catalyst lowers the activation energy to 55kJ/mo

notka56 [123]

Answer:

a) at 25°C the rate of reaction increases by a factor of 1,027*10^7

b) at 25°C the rate of reaction increases by a factor of 1,777*10^5

Explanation:

using the Arrhenius equation

k= ko*e^(-Ea/RT)

where

k= reaction rate

ko= collision factor

Ea= activation energy

R= ideal gas constant= 8.314 J/mol*K

T= absolute temperature

for the uncatalysed reaction

k1= ko*e^(-Ea1/RT)

for the catalysed reaction

k2= ko*e^(-Ea2/RT)

dividing both equations

k2/k1= e^(-(Ea2-Ea1)/RT)

a) at 25°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,027*10^7

therefore at 25°C , k2/k1 = 1,027*10^6

b) at 125°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,777*10^5

therefore at 125°C , k2/k1 = 1,777*10^5

Note:

when the catalysts is incorporated, the catalysed reaction and the uncatalysed one run in parallel and therefore the real reaction rate is

k real = k1 + k2 = k2 (1+k1/k2)

since k2>>k1 → 1+k1/k2 ≈ 1 and thus k real ≈ k2

6 0

3 years ago

Which compound is classified as a hydrocarbon?

Andrej [43]

<span>(2) butynu........................</span>

5 0

3 years ago

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