How many moles of O2 are consumed if 20 moles of SO2 are produced

if 28.5 g of calcium hydroxide is dissolved in enough water to make 185g of solution what is the percent by mass of calcium hydr

DENIUS [597]

The percent by mass of calcium hydroxide in the solution : 15.41%

<h3>Further explanation</h3>

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight/volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.

Mass of solute (Ca(OH₂-Calcium hydroxide) : 28.5

Mass of solution = 185 g

$\tt \%mass=\dfrac{mass~solute}{mass~solution}\times 100\%\\\\\%mass=\dfrac{28.5~g}{185~g}\times 100\%\\\\\%mass=15.41\%$

6 0

3 years ago

In the gas-phase reaction 2 A + B ⇌ 3 C + 2 D, it was found that, when 1.00 mol A, 2.00 mol B, and 1.00 mol D were mixed and all

Naddik [55]

Answer:

(i)The mole fractions are :

$A=\frac{0.4}{4.3} \\=0.0930$

$B=\frac{1.4}{4.3} \\=0.3256$

$C=\frac{0.9}{4.3} \\=0.2093$

$D=\frac{1.6}{4.3} \\=0.3721$

(ii) $K=0.4508$

(iii)ΔG = 1.974kJ

Explanation:

The given equation is :

$2A+B$ ⇄ $3C+2D$

Let $\alpha$ be the number of moles dissociated per mole of $B$

Thus ,

<em>The initial number of moles of :</em>

$A=1$

$B=2$

$C=0$

$D=1$

$2A\\(1-2\alpha)$ + $B\\2(1-\alpha)$ ⇄ $3C\\(3\alpha)$ + $2D\\(1+2\alpha)$

And finally the number of moles of $C[tex] is 0.9Thus ,[tex]3\alpha=0.9\\\alpha=0.3[tex]The final number of moles of:[tex]A = 1-2\alpha=1-2*0.3=0.4mol[tex] [tex]B=2(1-\alpha)=2(1-0.3)=1.4mol[tex][tex]D=1+2\alpha=1+2*0.3=1.6mol[tex]Thus , total number of moles are : 0.4+1.4+0.9+1.6=4.3(i)The mole fractions are : [tex]A=\frac{0.4}{4.3} \\=0.0930$

$B=\frac{1.4}{4.3} \\=0.3256$

$C=\frac{0.9}{4.3} \\=0.2093$

$D=\frac{1.6}{4.3} \\=0.3721$

(ii)

$K=\frac{(P_C^3)(P_D^2)}{(P_A^2)(P_B)}$

Where ,

$P_A,P_B,P_C,P_D$ are the partial pressures of A,B,C,D respectively.

Total pressure = 1 bar .

∴

$K=\frac{0.2093^3*0.3721^2}{0.0930^2*0.3256} \\K=0.4508$

(iii)

Δ $G=-RTlnK\\$

ΔG = $-8.314*(273+25)*ln(0.4508)\\=1973.96J\\=1.974kJ$

3 0

3 years ago

The element magnesium is a

kipiarov [429]

Group 2
Period 3
Alkaline Earth Metal

3 0

3 years ago

Helppppppp! why should i use spr??

Viktor [21]

SPR can be used in order to do real time monitoring and to evaluate the advancement of compounds.

<h3>Why should we use SPR?</h3>

SPR provide information to evaluate whether or not compounds should advance to the next stage of investigation. It also provides real-time monitoring.

So we can conclude that SPR can be used in order to do real time monitoring and to evaluate the advancement of compounds.

Learn more about monitoring here: brainly.com/question/13163394

#SPJ1

6 0

2 years ago

the distance between atomes is sometimes given in picometers,where 1pm is equivalent to 1×10^12m if the distance between the car

grigory [225]

Answer:

Distance between the carbon atom = 491 pm

Explanation:

Given:

1 pm = 1 × 10⁻¹² m

Distance between the carbon atom = 2.81 × 10⁻⁸ cm

Find:

Distance in picometers

Computation:

Distance between the carbon atom = 2.81 × 10⁻⁸ cm = 491 X 10⁻¹² m

Distance between the carbon atom = 491 pm

3 0

3 years ago