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olga nikolaevna [1]
2 years ago
4

Two glasses labeled A and B had equal amounts of water. Ted heated the water in one of the glasses. He then put two drops of foo

d coloring into each of the two glasses. The table shows the time taken by the food coloring in the two glasses to spread out. Food Coloring Experiment Glass Time A 25 seconds B 6 seconds Which of the following statements is correct? The water in Glass B was heated; therefore, the particles in Glass B have greater kinetic energy and move faster. The water in Glass A was heated; therefore, the particles in Glass A have greater kinetic energy and move faster. The water in Glass B was heated; therefore, the particles in Glass B move slower. The water in Glass A was heated; therefore, the particles in Glass A move slower.
Chemistry
1 answer:
Alik [6]2 years ago
5 0

The correct option is this: THE WATER IN GLASS B WAS HEATED; THEREFORE, THE PARTICLES IN GLASS B HAVE GREATER KINETIC ENERGY AND MOVE FASTER.

When heat is added to a liquid such as water, the heat energy transferred to the liquid usually weaken the attractive forces holding the water molecules together making it possible for the water molecules to overcome the chemical bonds and to move freely and faster with higher kinetic energy.

From the information given in the question, you can see that it takes only 6 seconds for the food coloring particles in glass B to travel round compare to glass A, which takes 25 seconds. This implies that, the water particles in glass A have been heated, they have higher kinetic energy and thus, they are able to carry the food coloring particles round the water in a very short time.

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34.62 mL of 0.1510 M NaOH was needed to neutralize 50.0 mL of an H2SO4 solution. What is the concentration of the original sulfu
Igoryamba

0.1045M is the concentration of the original sulfuric acid solution

Explanation:

Titration is done to know the volume or concentration of unknown electrolyte.

Data given:

volume of acid = 50 ml

molarity of acid =?

volume of base NaOH = 34.62 ml

molarity of the base = 0.1510

For titration the formula used is,

Macid x Vacid = Mbase  x V base

Putting the values in above equation:

M acid x 50 = 34.62 x 0.1510

Macid = \frac{5.22762}{50}

           = 0.1045 M is the molarity of sulphuric acid solution used in neutralization of 0.1510 M base solution.

5 0
2 years ago
Which equation shows the proper setup to find the volume of a 0.250 M CaCl2 solution that contains 39.5 g of CaCl2?
Jlenok [28]
Molarity = number of mole of substance(n) / volume of solution (V).

n(CaCl2) = mass (CaCl2)/M(CaCl2)

M(CaCl2) = 40+2*35.5 = 111 g/mol 

n(CaCl2) =39.5 g CaCl2*1 mol/111g

0.250 M = 39.5 g CaCl2*1 mol/111g*volume of solution (V).

volume of solution (V) = 39.5 g CaCl2*1 mol/(0.250 M*111g) = 1.42 L
4 0
2 years ago
Read 2 more answers
ethylene glycol used in automobile antifreeze and in the production of polyester. The name glycol stems from the sweet taste of
Luden [163]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is CH_3O and C_4H_{12}O_4

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=9.06g

Mass of H_2O=5.58g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 9.06 g of carbon dioxide, \frac{12}{44}\times 9.06=2.47g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 5.58 g of water, \frac{2}{18}\times 5.58=0.62g of hydrogen will be contained.

  • Mass of oxygen in the compound = (6.38) - (2.47 + 0.62) = 3.29 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.206 moles.

For Carbon = \frac{0.206}{0.206}=1

For Hydrogen  = \frac{0.62}{0.206}=3

For Oxygen  = \frac{0.206}{0.206}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 3 : 1

Hence, the empirical formula for the given compound is C_1H_{3}O_1=CH_3O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 124 amu = 124 g/mol

Mass of empirical formula = 31 g/mol

Putting values in above equation, we get:

n=\frac{124g/mol}{31g/mol}=4

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 4)}H_{(3\times 4)}O_{(1\times 4)}=C_4H_{12}O_4

Thus, the empirical and molecular formula for the given organic compound is CH_3O and C_4H_{12}O_4

3 0
2 years ago
What is the mass of 1.82 moles of lithium carbonate?
Svetlanka [38]
You can view more details on each measurement unit: molecular weight of Lithium Carbonate or grams The molecular formula for Lithium Carbonate is Li2CO3. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles Lithium Carbonate, or 73.8909 grams.
5 0
2 years ago
Read 2 more answers
The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s
denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0
2 years ago
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