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ivolga24 [154]
3 years ago
10

Trevor dropped a bit of olive oil on the kitchen floor while he was cooking. Later, his mother slipped on the spot where he drop

ped the oil and fell.
What can Trevor infer based on his observations?
A.
Trevor's mother slipped on the spot where he dropped the oil.
B.
Olive oil can be used as an adhesive between two solid objects.
C.
The oil reduced the friction between the floor and his mother's foot.
D.
The oil increased the gravitational force on his mother's foot.
Chemistry
2 answers:
anastassius [24]3 years ago
8 0
The correct answer is C
Korolek [52]3 years ago
5 0

Answer:

C.

The oil reduced the friction between the floor and his mother's foot. Im not 100% sure though! :)

Explanation:

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Using the model of the periodic table, which two elements pictured have similar chemical properties?
Murljashka [212]

Answer:

1 and 3

Explanation:

The vertical columns (groups) of the periodic table are arranged such that all its elements have the same number of valence electrons. All elements within a certain group thus share similar properties.

5 0
3 years ago
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Hydrogen reacts with chlorine to form hydrogen chloride (HCl (g), mc012-1.jpgHf = –92.3 kJ/mol) according to the reaction below.
Kobotan [32]

<u>Given:</u>

H2(g) + Cl2 (g) → 2HCl (g)

<u>To determine:</u>

The enthalpy of the reaction and whether it is endo or exothermic

<u>Explanation:</u>

Enthalpy of a reaction is given by the difference between the enthalpy of formation of reactants and products

ΔH = ∑nHf (products) - ∑nHf (reactants)

      = [2Hf(HCl)] - [Hf(H2) + Hf(Cl2)] = 2 (-92.3) kJ = - 184.6 kJ

Since the reaction enthalpy is negative, the reaction is exothermic

<u>Ans:</u> The enthalpy of reaction is -184. kJ and the reaction is exothermic

3 0
3 years ago
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4. Explain what you’ve learned about significant figures. How many significant figures are in the measurement 0.03050 kg?
nasty-shy [4]

There are 4 significant figures! Start counting after the first non-zero digit :)

Hope this helps.

3 0
3 years ago
A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2
a_sh-v [17]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is CHO_2 and C_2H_2O_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=13.00g

Mass of H_2O=2.662g

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

So, in 13.00 g of carbon dioxide, \frac{12}{44}\times 13.00=3.54g of carbon will be contained.

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

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

So, in 2.662 g of water, \frac{2}{18}\times 2.662=0.296g of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 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{3.54g}{12g/mole}=0.295moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles

  • <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 0.295 moles.

For Carbon = \frac{0.295}{0.295}=1

For Hydrogen = \frac{0.296}{0.295}=1

For Oxygen = \frac{0.603}{0.295}=2.044\approx 2

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

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

Hence, the empirical formula for the given compound is CHO_2

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 = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

n=\frac{90.04g/mol}{45g/mol}=2

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

C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4

Hence, the empirical and molecular formula for the given organic compound is CHO_2 and C_2H_2O_4

3 0
3 years ago
a solution is made by completely dissolving 90. grams of kno3 in 100 grams of water in a beaker. the temperature of this solutio
jolli1 [7]
Pressure has no effect on the solubility of KNO3 in water. This is because it is solid in liquid type of solution. In solid in liquid type of solution, solid is solute (minor component), liquid is solvent (major component). For solid in liquid type of solutions, solubility is independent of pressure. 

On other hand, pressure has a pronounced effect on the solubility of gas in liquid type solutions. In such system, gas is solute (minor component) and liquid is solvent (major component). Example of such solution is aerated water. Herein, CO2 is dissolved in water. In such gas in liquid type of solutions, solubility increases with increasing pressure. 
3 0
3 years ago
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