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2 years ago

Webb has calculated the percent composition of a compound. How can he check his result?

1 answer:

7 0

Webb has calculated the percent composition of a compound. He can check his result by adding them to see if they equal up to 100. Why? Well, percent composition tells the quantity of elements with 100 as a base of total amount. This means that it will have to add to 100 to check the result. You would add up all of the values of percent composition of elements to see if they equal 100, and if they do, the results are accurate.

Your final answer: Webb can check his result by seeing if they add up to 100, considering that is the base total quantity.

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Select the correct answer.

Minchanka [31]

Answer:

A. it is the lowest at low temperatures

Explanation:

It is true with respect to the kinetic energy of a molecule that the it is the lowest at low temperatures.

The kinetic energy of a molecule is the energy due to the motion of the particles within a substance.

Kinetic energy is directly proportional to the temperature of a substance.
The higher the temperature, the more the kinetic energy of the molecules within a system.
At low temperature, kinetic energy is the lowest.
At the highest temperature, kinetic energy is the highest

3 0

3 years ago

Read 2 more answers

Which or these equations is balanced H2SO4 + 2Al > Al2(SO4)3 + H2 or 2KCl + Pb(NO3)2 > 2KNO3 + PbCl2

Dmitry_Shevchenko [17]

Balancing of chemical equation is essential because of the law of conservation of mass, which states that the mass of a system can not be created or removed.

The second equation is balanced

$2KCl_(_a_q_) +Pb(NO_3)_2_(_a_q_) ==> 2KNO_3_(-a_q_) + PbCl_2(_a_q_)$

This is because the number of elements of each atom in the product side equal the number of elements of each atom on the reactant side.

The first equation is not balanced

$H_2SO_4 _(_a_q) + 2Al_(_s_) ==> Al(SO_4)_3_(_a_q_) + H_2_(_g_)$

This is because there is 1 molecule of $SO_4^{-2}$ on reactant side as compared to 3 molecules of $SO_4^{-2}$

To balance the equation we add a coefficient of 3 on sulphuric acid ( $H_2SO_4$ ) and a coefficient of 3 on hydrogen ( $H_2$ )

$3H_2SO_4_(_a_q_) + 2Al_(_s_) ==> Al(SO_4)_3_(_a_q_) + 3H_2_(_g_)$

7 0

4 years ago

A 6.165 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 10.27 grams of CO2 and 3

e-lub [12.9K]

Answer:

Explanation:

mass of carbon in 10.27 g of CO₂ = 12 x 10.27 / 44 = 2.80 g

mass of hydrogen ( H ) in 3.363 g of H₂O = 2 x 3.363 / 18

= .373 g

These masses would have come from the sample of 6.165 g .

Rest of 6.165 g of sample is oxygen .

So oxygen in the sample = 6.165 - ( 2.8 + .373 ) = 2.992 g

Ratio of C , H , O in the sample

2.8 : .373 : 2.992

C: H : O : : 2.8 : .373 : 2.992

Ratio of moles

C: H : O : : 2.8/12 : .373/1 : 2.992 / 16

C: H : O : : .2333 : .373 : .187

C: H : O : : .2333/.187 : .373/.187 : .187/.187

C: H : O : : 1.247 : 1.99 : 1

C: H : O : : 5 : 8 : 4 ( after multiplying by 4 )

Hence empirical formula

C₅H₈O₄

Molecular formula ( C₅H₈O₄ )n

n ( 5 x 12 + 8 x 1 + 4 x 16 ) = 132

n x ( 60 + 8 + 64 ) = 132

n = 1

Molecular formula = C₅H₈O₄.

3 0

3 years ago

Which of the following are balanced equations? Check all that apply.

EleoNora [17]

Answer: the answer is b

Explanation:

because why not

8 0

3 years ago

Calculate the maximum volume in ml of 0.15M HCl that each of the following antacid formulations would be expected to neutralize.

vlada-n [284]

a. 34 mL; b. 110 mL

a. A tablet containing 150 Mg(OH)₂

Mg(OH)₂ + 2HCl ⟶ MgCl₂ + 2H₂O

Moles of Mg(OH)₂ = 150 mg Mg(OH)₂ × [1 mmol Mg(OH)₂/58.32 mg Mg(OH)₂

= 2.572 mmol Mg(OH)₂

Moles of HCl = 2.572 mmol Mg(OH)₂ × [2 mmol HCl/1 mmol Mg(OH)₂]

= 5.144 mmol HCl

Volume of HCl = 5.144 mmol HCl × (1 mmol HCl/0.15 mmol HCl) = 34 mL HCl

b. A tablet containing 850 mg CaCO₃

CaCO₃ + 2HCl ⟶ CaCl₂ + CO₂ + H₂O

Moles of CaCO₃ = 850 mg CaCO₃ × [1 mmol CaCO₃/100.09 mg CaCO₃

= 8.492 mmol CaCO₃

Moles of HCl = 8.492 mmol CaCO₃ × [2 mmol HCl/1 mmol CaCO₃]

= 16.98 mmol HCl

Volume of HCl = 16.98 mmol HCl × (1 mL HCl/0.15 mmol HCl) = 110 mL HCl

5 0

3 years ago