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Artist 52 [7]
3 years ago
10

An antacid tablet weighing 0.833 g contained calcium carbonate as the active ingredient, in addition to an inert binder. When an

acid solution weighing 58.072 g was added to the tablet, carbon dioxide gas was released, producing a fizz. The resulting solution weighed 57.053 g. How many grams (g) of carbon dioxide were produced?
Chemistry
1 answer:
Naddika [18.5K]3 years ago
3 0

Answer:

1.852 g of CO2 were produced in the chemical reaction

Explanation:

The problem is pretty simple. We can write down the chemical reaction that is involved to help us understand better what is going on in the process:

CaCO3 (aq) + HAc (aq) → CaAc (aq) + CO2 (g) + H2O (l)

Let's think this through: we have a tablet that has an active compound (CaCO3) and an inert substance that weighs 0.833 g. When we add 58.072 g of an acid solution (represented in the equation as HAc because we are not told specifically which acid is being added), CO2 is formed and released from solution as gas leaving us with an aqueous solution that weighs 57.053 g.

Having said that, we know that the only mass lost during the reaction is due to the formation of CO2 gas. Therefore, we sum the reactants (the tablet + the acid solution) and subtract the mass of the remnant solution. This value will indicate us the amount of CO2 formed:

0.833 g of the Tablet + 58.072 g from the Acid solution = 58.905 g

58.905 g of reactants - 57.053 g of remnant solution = 1.852 g of produced CO2.

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3 0
3 years ago
Draw an ether that contains exactly five carbon atoms and only single bonds
Eddi Din [679]

<em>Answer:</em>

  •                                       H3CH2COCH2CH2CH3

<em>Ether:</em>

Ether are organic compounds that contain ether functional group , in which oxygen atom is connected with two alkyl or aryl group.

They have general formula as follow

  • R---O---R   or R'---O----R or R'---O---R'

            while  R = Alkyl

                       R' = Aryl

8 0
3 years ago
A gas sample with a volume of 1,500 cm® is heated from -65 °C to 75 °C. Assuming the
djverab [1.8K]

Answer:

V₂ = 2509.62 cm³

Explanation:

Given data:

Initial volume = 1500 cm³

Initial temperature = -65°C (-65 + 273 = 208 K)

Final temperature = 75°C ( 75 +273 = 348 K)

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 1500 cm³ × 348 K / 208 k

V₂ = 522000 cm³.K / 208 k

V₂ = 2509.62 cm³

7 0
3 years ago
Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp
gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200  

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in  j/mol R Ea is

Ea=35.5*1000 j/mol R

ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

Read more about rate constant

brainly.com/question/20305871

#SPJ1

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Answer: O2+6H12O6=CO2+ENERGY(ATP)

I DON'T THINK SHE IS CORRECT

Explanation:

5 0
3 years ago
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