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kotegsom [21]
3 years ago
7

For the reaction, calculate how many grams of the product form when 1.8 g of cao completely reacts. assume that there is more th

an enough of the other reactant. cao(s)+co2(g)→caco3(s)
Chemistry
1 answer:
ludmilkaskok [199]3 years ago
5 0
Balance Chemical Equation is,

                            <span>CaO </span>₍s₎  <span>+  CO</span>₂ ₍g₎   <span>→   CaCO</span>₃ ₍s₎

According to Equation

56.07 g of CaO when reacted with excess CO₂ produces  =  100.08 g of CaCO₃.

Then,

    1.8 g of CaO is reacted with excess CO₂  =  X g of CaCO₃ will produce.

Solving for X,
                               X  =  (100.08 g × 1.8 g) ÷ 56. 07 g

                               X  =  3.21 g of CaCO₃
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Given the formula: what is a chemical name of this compound? propane propanal propanol propanone
katrin [286]
1) Chemical formula for propane is CH₃-CH₂-CH₃. 
Propane is a three carbon alkane (acyclic saturated <span>hydrocarbon).
</span>2) Chemical formula for propanal is CH₃-CH₂-CH=O.
Propanal <span> is a </span>saturated<span> three carbon </span>aldehyde (have<span> a </span>carbonyl<span> center).
3) </span>Chemical formula for propanol is CH₃-CH₂-CH₂-OH.
1-propanol <span> is a </span><span>primary alcohol.
4) </span>Chemical formula for propanone is (CH₃)₂-C=O.
Propanone or acetone is <span>he simplest and smallest</span> ketone.
7 0
3 years ago
A change in matter that produces one or more new substances is a(n)
lukranit [14]

You could use another word for change can be variable witch means change and if you times the one two more times then you would get four because two time two would be four and times the one would be four.


6 0
3 years ago
The volume of air in a person’s lungs is 615 mL at a pressure of 760. mmHg. Inhalation occurs as the pressure in the lungs drops
chubhunter [2.5K]

<u>Answer:</u> The final volume of lungs is 621.5 mL

<u>Explanation:</u>

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.

The equation given by this law is:

P_1V_1=P_2V_2

where,

P_1\text{ and }V_1 are initial pressure and volume.

P_2\text{ and }V_2 are final pressure and volume.

We are given:

P_1=760mmHg\\V_1=615mL\\P_2=752mmHg\\V_2=?mL

Putting values in above equation, we get:

760mmHg\times 615mL=752mmHg\times V_2\\\\V_2=\frac{760\times 615}{752}=621.5mL

Hence, the final volume of lungs is 621.5 mL

8 0
3 years ago
What is the molarity of a solution that contains 1000.0 mg of AgNO3 that has been dissolved in 500 mL of water
Vaselesa [24]

0.012moldm⁻³

Explanation:

Given parameters:

Mass of AgNO₃  = 1000mg

Volume of water = 500mL

Unknown:

Molarity of solution  = ?

Solution:

The molarity of a solution is the number of moles of a solute dissolved in volume of solvent.

 Molarity = \frac{xnumber of moles}{Volume}

 

Number of moles of AgNO₃  = ?

   Number of moles = \frac{mass}{molar mass}

Molar mass of AgNO₃ = 108 + 14 + 3(16) = 170g/mol

   convert mass to g;

      1000mg = 1g

 Number of moles  = \frac{1}{170}  = 0.00588moles

   convert the given volume to dm³;

       1000mL  = 1dm³;

        500mL = 0.5dm³

Now solve;

  Molarity = \frac{0.00588}{0.5}  = 0.012moldm⁻³

learn more:

Molarity brainly.com/question/9324116

#learnwithBrainly

4 0
3 years ago
Calculate the pH of the following simple solutions:
IceJOKER [234]

Answer :

(1) pH = 1.27

(2) pH = 13.35

(3) The given solution is not a buffer.

Explanation :

<u>(1) 53.1 mM HCl</u>

Concentration of HCl = 53.1mM=53.1\times 10^{-3}M

As HCl is a strong acid. So, it dissociates completely to give hydrogen ion and chloride ion.

So, Concentration of hydrogen ion= 53.1\times 10^{-3}M

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

pH=-\log [H^+]

pH=-\log (53.1\times 10^{-3})

pH=1.27

<u>(2) 0.223 M KOH</u>

Concentration of KOH = 0.223 M

As KOH is a strong base. So, it dissociates completely to give hydroxide ion and potassium ion.

So, Concentration of hydroxide ion= 0.223 M

Now we have to calculate the pOH.

pOH=-\log [OH^-]

pOH=-\log (0.223)

pOH=0.65

Now we have to calculate the pH.

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-0.65=13.35

<u>(3) 53.1 mM HCl + 0.223 M KOH</u>

Buffer : It is defined as a solution that maintain the pH of the solution by adding the small amount of acid or a base.

It is not a buffer because HCl is a strong acid and KOH is a strong base. Both dissociates completely.

As we know that the pH of strong acid and strong base solution is always 7.

So, the given solution is not a buffer.

5 0
3 years ago
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