CHEMISTRY COLLEGE

Which process is used to make lime (calcium oxide) from limestone (calcium carbonate)?

Answers

Answer 1

Answer:

Answer:

Explanation:

Calcium oxide is fromed by the decomopostion of CaCO3 at high temperature.

CaCO3 ------> CaO +CO2

Hope this helps you

Related Questions

In the laboratory a "coffee cup" calorimeter, or constant pressure calorimeter, is frequently used to determine the specific heat of a solid, or to measure the energy of a solution phase reaction. A student heats 62.08 grams of magnesium to 97.96 °C and then drops it into a cup containing 77.81 grams of water at 23.19 °C. She measures the final temperature to be 35.60 °C. The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.79 J/°C. Assuming that no heat is lost to the surroundings calculate the specific heat of magnesium.
Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical formulas. Calculate the empirical formulas of the following polymers: (a) Lucite (Plexiglas); 59.9% C, 8.06% H, 32.0% O (b) Saran; 24.8% C, 2.0% H, 73.1% Cl (c) polyethylene; 86% C, 14% H (d) polystyrene; 92.3% C, 7.7% H (e) Orlon; 67.9% C, 5.70% H, 26.4% N
Most wine is prepared by the fermentation of the glucose in grape juice by yeast: C6H12O6(aq) --> 2C2H5OH(aq) + 2CO2(g) How many grams of glucose should there be in grape juice to produce 725 mLs of wine that is 11.0% ethyl alcohol, C2H5OH (d=0.789 g/cm3), by volume?
A 0.1153-gram sample of a pure hydrocarbon was burned in a C-H combustion train to produce 0.3986 gram of CO2and 0.0578 gram of H2O. Determine the masses of C and H in the sample and the percentages of these elements in this hydrocarbon.
A weather balloon has a volume of 200.0 L at a pressure of 760 mm Hg. As it rises, the pressure decreases to 282 mm Hg. What is the new volume of the balloon? (Assume constant temperature)

Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it is used to make the refrigerant carbon tetrafluoride, CF4 via the reaction 2COF2(g)⇌CO2(g)+CF4(g), Kc=4.90 If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?

Answers

Answer : The concentration of $COF_2$ remains at equilibrium will be, 0.37 M

Explanation : Given,

Equilibrium constant = 4.90

Initial concentration of $COF_2$ = 2.00 M

The balanced equilibrium reaction is,

$2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)$

Initial conc. 2 M 0 0

At eqm. (2-2x) M x M x M

The expression of equilibrium constant for the reaction will be:

$K_c=([CO_2][CF_4])/([COF_2]^2)$

Now put all the values in this expression, we get :

$4.90=((x)* (x))/((2-2x)^2)$

By solving the term 'x' by quadratic equation, we get two value of 'x'.

$x=1.291M\text{ and }0.815M$

Now put the values of 'x' in concentration of $COF_2$ remains at equilibrium.

Concentration of $COF_2$ remains at equilibrium = $(2-2x)M=[2-2(1.219)]M=-0.582M$

Concentration of $COF_2$ remains at equilibrium = $(2-2x)M=[2-2(0.815)]M=0.37M$

From this we conclude that, the amount of substance can not be negative at equilibrium. So, the value of 'x' which is equal to 1.291 M is not considered.

Therefore, the concentration of $COF_2$ remains at equilibrium will be, 0.37 M

Find the weighted average of these values. Col1 Value 5.00 6.00 7.00Col2 Weight 75.0 % 15.0 % 10.0 %

Answers

The weighted average (Avg) for these values has been 5.35.

The weighted average has been an arithmetic calculation of the mean value for the percent abundance of each value.

Computation for weighted average

The weighted average (Avg) for the values has been given by:

$Avg=V_1\;*\;(W_1)/(100)\;+\;V_2\;*\;(W_2)/(100)\;+\;V_3\;*\;(W_3)/(100)$

The values have been given,

$V_1=5\nV_2=6\nV_3=7$

The weighted average has been given as:

$W_1=75\nW_2=15\nW_3=10$

For the given set of values, the weighted average (Avg) has been given as:

$Avg=5\;*\;(75)/(100)\;+\;6\;*\;(15)/(100)\;+\;7\;*\;(10)/(100)\nAvg=5\;*\;0.75\;+\;6\;*\;.015\;+\;7\;*\;0.1\nAvg=5.35$

The weighted average (Avg) for these values has been 5.35.

Learn more about weighted average, here:

brainly.com/question/18554478

Answer:

5.35

Explanation:

Value 5.00 6.00 7.00

Weight 75.0% 15.0% 10.0 %

We can determine the weighted average of these values using the following expression.

Weighted average = ∑ wi × xi

where,

w: relative weight

x: value

Weighted average = 5.00 × 0.750 + 6.00 × 0.150 + 7.00 × 0.100

Weighted average = 5.35

wo reactions and their equilibrium constants are given. A + 2 B − ⇀ ↽ − 2 C K 1 = 2.57 2 C − ⇀ ↽ − D K 2 = 0.226 A+2B↽−−⇀2CK1=2.572C↽−−⇀DK2=0.226 Calculate the value of the equilibrium constant for the reaction D − ⇀ ↽ − A + 2 B .

Answers

Answer: The value of equilibrium constant for the net reaction is 11.37

Explanation:

The given chemical equations follows:

Equation 1: $A+2B\xrightarrow[]{K_1} 2C$

Equation 2: $2C\xrightarrow[]{K_2} D$

The net equation follows:

$D\xrightarrow[]{K} A+2B$

As, the net reaction is the result of the addition of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K=K_1* (1)/(K_2)$

We are given:

$K_1=2.57$

$K_2=0.226$

Putting values in above equation, we get:

$K=2.57* (1)/(0.226)=11.37$

Hence, the value of equilibrium constant for the net reaction is 11.37

Calculate the concentration of OH in a solution that contains 3910-4 M H30 at 25°C. Identify the solution as acidic, basic or neutral OA) 2.6 10-11 M, acidic OB)26 10-11 M. basic O c) 3.9 x 10-4 M, neutral OD) 2.7 * 10-2 M

Answers

Answer : The correct option is, (A) $2.6* 10^(-11)M$ , acidic

Explanation:

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

When the value of pH is less then 7 then the solution will be acidic.

When the value of pH is more then 7 then the solution will be basic.

When the value of pH is equal to 7 then the solution will be neutral.

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (3.9* 10^(-4))$

$pH=3.41$

Now we have to calculate the pOH.

$pH+pOH=14\n\npOH=14-pH\n\npOH=14-3.41=10.59$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$10.59=-\log [OH^-]$

$[OH^-]=2.6* 10^(-11)M$

Therefore, the $OH^-$ concentration is, $2.6* 10^(-11)M$

QUICK WILL MARK BRAINLIEST!!Match each of the unknown ions to its appropriate description.

A− A) A nonmetal that gained one electron
B+ B) A metal that lost one electron
C2− C) A metal that lost two electrons
D2+ D) A nonmetal that gained two electrons

Answers

Answer:

c2-c

Explanation:

A mental that lost two electrons just joking I don't k ow but I think it ain't the first o e because of comment sense to the 5th power

Answer:

A and B

Explanation:

metal form ion by lossing elecron

non metal form ion by gaining electron

The federal limit for cadmium in drinking water is 0.010 mg per liter of solution. What is the molar concentration of a Cd solution that has reached the limit?

Answers

Thank you for posting your question here. The molar concentration of a Cd solution that has reached the limit is 8.89x10^-8 mols/L. The equation to be used M=n/L to solve the above problem. Below is the solution:

.010mg/L = .00001g/L
.00001g / 112.41g/mol = 8.89x10^-8 mols/L

Answer:

$M=8.9x10^(-8)M$

Explanation:

Hello,

In this case, one can assume 1L as the volume of the solution, so we've got 0.010mg of cadmium. Now, as we're asked to know its molarity, one computes the moles of cadmium as follows:

$n_(Cd)=0.010mg*(1x10^(-3)gCd)/(1mgCd)*(1molCd)/(112.4gCd)=8.9x10^(-8)molCd$