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olga_2 [115]
3 years ago
11

If a car is traveling on the highway at a constant velocity, the force that pushes the car forward must be

Chemistry
2 answers:
Slav-nsk [51]3 years ago
5 0
<span>If a car is traveling on the highway at a constant velocity, the force that pushes the car forward must be
</span>acceleration

Mandarinka [93]3 years ago
5 0
Well, when your tires and the road are mixed together at a high speed, it causes a friction force, which is what propels the car forward, even when you are not using the gas peddle. 
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What was one main point of Dalton’s atomic theory
s344n2d4d5 [400]

Answer:

Everything is composed of atoms! which are the indivisible building blocks of matter and cannot be destroyed. All atoms of an element are identical. The atoms of different elements vary in size and mass.

Explanation:

Hope this helped

8 0
3 years ago
Read 2 more answers
A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO
Akimi4 [234]

Answer: The empirical formula for the given compound is CH_2

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of CO_2=22.0g

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

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

So, in 22.0 g of carbon dioxide, \frac{12}{44}\times 22.0=6g of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

Mass of hydrogen = 7.0 g - 6 g

Mass of hydrogen = 1.0 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles

Step 2: 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 0.5 moles.

For Carbon = \frac{0.5}{0.5}=1

For Hydrogen  = \frac{1.0}{0.5}=2

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is C_{1}H_{2}=CH_2

4 0
4 years ago
According to naming rules, the types of compound that use prefixes in their names are ________.
arlik [135]

Answer:

Covalent compounds.

Explanation:

Hello,

In this case, when forming chemical bonds in order to form compounds, we say that if electrons are shared, covalent compounds are to be formed and they usually have subscripts that need prefixes to be named, for instance phosphorous pentachloride (PCl5), dichlorine heptoxide (Cl2O7), carbon tetrachloride (CCl4) and many others.

Regards.

6 0
3 years ago
Which pair of atoms has the highest electronegativity difference?
Tems11 [23]
The answer its O-H.........
7 0
3 years ago
Read 2 more answers
Determine the empirical formula of a compound containing 48.38 grams of carbon, 6.74 grams of hydrogen, and 53.5 grams of oxygen
Darya [45]

Answer: The empirical formula of the compound becomes CH_2O

<u>Explanation:</u>

The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

We are given:

Mass of C = 48.38 g

Mass of H = 6.74 g

Mass of O = 53.5 g

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

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

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

Molar mass of C = 12 g/mol

Molar mass of H = 1 g/mol

Molar mass of O = 16 g/mol

Putting values in equation 1, we get:

\text{Moles of C}=\frac{48.38g}{12g/mol}=3.023 mol

\text{Moles of H}=\frac{6.74g}{1g/mol}=6.74 mol

\text{Moles of O}=\frac{53.5g}{1g/mol}=3.34 mol

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

Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 3.023 moles

\text{Mole fraction of C}=\frac{3.023}{3.023}=1

\text{Mole fraction of H}=\frac{6.74}{3.023}=2.23\approx 2

\text{Mole fraction of O}=\frac{3.34}{3.023}=1.105\approx 1

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

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

Hence, the empirical formula of the compound becomes CH_2O

5 0
3 years ago
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