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gladu [14]
2 years ago
7

Research Question: Does the type of gasoline put in a car effect how fast the car can drive?

Chemistry
1 answer:
Jobisdone [24]2 years ago
6 0

Answer:

type of gasoline

Explanation:

Independent variable in an experiment is the variable that the experimenter changes or manipulates in order to bring out a measurable outcome or response.

In this experiment involving how the type of gasoline put in a car affect how fast the car can drive, the independent variable is the TYPE OF GASOLINE because it is what the experimenter changes in order to see it's effect on the speed of the car (dependent variable).

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A 100.0 mL sample of a 0.200 M aqueous solution of K2CrO4 was added to 100.0 mL of a 0.100 M aqueous solution of BaCl2. The mixt
stepan [7]
The correct answer is 2.53 g of precipitate, BaCrO4.

5 0
2 years ago
sp2 hybrid orbitals have 1. trigonal bipyramidal symmetry. 2. linear symmetry. 3. tetrahedral symmetry. 4. trigonal planar symme
Cloud [144]

Answer:

4. trigonal planar symmetry.

Explanation:

The sp2 hybridization is formed with one s and two p atomic orbitals and form  trigonal planar symmetry. In sp2 hybridization, there are same valence shell in both the orbitals and it gives three equivalent sp2 hybridized orbitals that are separated by 120 degrees giving trigonal planar symmetry.

Hence, the correct answer is "4. trigonal planar symmetry.".

7 0
3 years ago
When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular
Luda [366]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is C_2H_2O_4 and C_6H_4O_2

<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=1.03g

Mass of H_2O=0.14g

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 1.03 g of carbon dioxide, \frac{12}{44}\times 1.03=0.28g 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 0.14 g of water, \frac{2}{18}\times 0.14=0.016g of hydrogen will be contained.

  • Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 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{0.28g}{12g/mole}=0.023moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles

  • <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 0.00775 moles.

For Carbon = \frac{0.023}{0.00775}=2.96\approx 3

For Hydrogen  = \frac{0.016}{0.00775}=2.06\approx 2

For Oxygen  = \frac{0.00775}{0.00775}=1

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

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is C_3H_{2}O_1=C_3H_2O

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 = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

n=\frac{108.10g/mol}{54g/mol}=2

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

C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2

Thus, the empirical and molecular formula for the given organic compound is C_3H_2O and C_6H_4O_2

6 0
3 years ago
Nuclear power plants produce energy using fission. One common fuel, uranium-235, produces energy through the fission reaction 23
Vikentia [17]

Answer:

mass of U-235  = 15.9 g (3 sig. figures)

Explanation:

1 atom can produce -------------------------> 3.20 x 10^-11 J energy

x atoms can produce ----------------------> 1.30 x 10^12 J energy

x = 1.30 x 10^12 / 3.20 x 10^-11

x = 4.06 x 10^22 atoms

1 mol ----------------------> 6.023 x 10^23 atoms

y mol ----------------------> 4.06 x 10^22 atoms

y = 0.0675 moles

mass of U-235 = 0.0675 x 235 = 15.8625

mass of U-235  = 15.9 g (3 sig. figures)

7 0
3 years ago
How does a magnet separate mixtures like sand and iron filings?​
algol13

Answer:

because iron is magnetic and sand is not

Explanation:

6 0
3 years ago
Read 2 more answers
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