CHEMISTRY COLLEGE

What pressure is exerted on the bottom of a 0.500-m-wide by 0.900-m-long gas tank that can hold 50.0 kg of gasoline by the weight of the gasoline in it when it is full?

Answers

Answer 1

Answer:

Answer:

1088.89 Pa

Explanation:

According to the Newton's second law of motion:-

$Force=Mass* Acceleration$

Mass = 50.0 kg

Acceleration = g = 9.81 m/s²

So,

$Force=50.0* 9.8\ kgm/s^2$

Force = 490 N

Area of the base = $length* breath$ = $0.500* 0.900$ m² = 0.45 m²

Pressure = Force/Area = $(490\ N)/(0.45\ m^2)$ = 1088.89 Pa

Related Questions

g The electronic structure of which ONE of the following species cannot be adequately described by a single Lewis formula? (In other words, the electronic structure of which one can only be described by drawing two or more resonance structures?) A) C2H4 B) SO3 2– C) SO3 D) C3H8 E) HCN
If the enthalpy change of a reaction in a flask is δh = +67 kj, what can be said about the reaction?
The density of platinum is 21.45 g/cm3. What is the volume of a platinum sample with a mass of 11.2 g?
How many moles of MgS2O3 are in 223 g of the compound
A primary amine contains a five-membered carbon ring. The degree of unsaturation for the five-membered carbon ring is equal to three. In the chemical formula, there are seven carbon atoms. Determine its chemical formula.

How many liters of the antifreeze ethylene glycol [CH2(OH)CH2(OH)] would you add to a car radiator containing 6.50 L of water if the coldest winter temperature in your area is -10.°C? (The density of ethylene glycol is 1.11 g/mL. Assume the density of water at -10.°C is 1.00 g/mL.)

Answers

Answer:

Around 2.0 L of ethylene glycol needs to be added to the car radiator

Explanation:

The depression in freezing point ΔTf of a solution is directly proportional to its molality (m), i.e.

$\Delta T_(f)= T_(f)^(0)-T_(f)=i*K_(f)*m$

From the given information:

$T_(f)$ = freezing pt of solution = -10.0 C

$T_(f)^(0)$ = freezing pt of pure solvent = 0 C

Kf = freezing pt depression constant = 1.86 C/m

i = 1 for ethylene glycol antifreeze

$[0-(-10.0)] C= 1*(1.86 C/m) *( m)\n\nm = 5.38$

$Molality = (moles\ of\ solute)/(kg\ solvent) \n\nTherefore,\ moles of antifreeze = molality* mass\ of\ water\n$

$Molality = (moles\ of\ solute)/(kg\ solvent) \n\nTherefore,\ moles of ethylene glycol = molality* mass\ of\ water\n$

Volume of water = 6.50 L = 6500 ml

Density of water = 1.00 g/ml

Therefore mass of water = $density * volume = 1.00g/ml*6500ml = 6500g = 6.50kg$

$moles\ of\ ethylene glycol= 5.38moles/kg*6.50kg = 34.9 moles$

Molar mass of ethylene glycol = 62 g/mol

Mass of ethylene glycol needed = $molar\ mass* moles = 62g/mol*34.9moles=2163.8g$

Density of ethylene = 1.11 g/ml

Therefore, volume needed = $(mass)/(density) =(2163.8g)/(1.11g/ml) =1949ml$

16 = m x 1.86
m = 8.60 = moles solute / 6.50 Kg

moles solute = 55.9

mass solute = 55.9 x 62.068 g/mol=3470 g

V = 3479/ 1.11 =3126 mL= 3.13 L

delta T = 8.60 x 0.512 =4.40
boling point = 104.4 °C

Hope this helps.

The rate constant for a certain reaction is k = 6.50×10−3 s−1 . If the initial reactant concentration was 0.600 M, what will the concentration be after 3.00 minutes?

Answers

Answer: 1.037M

Explanation:

Since the rate constant unit is per seconds, therefore it is a first order reaction.

First order reaction equation is given as

InA= -kt +InAo

Where,Ao is the initial concentration of reactant =0.600M

A is the concentration of reactant at a specifies time t=3×60=180s

and k is the rate constant

InA = -6.50×10^-3 ×180 +In(0.6)

InA = -1.17 + 0.5108

InA= -1.680

A = e-1.680

A= 1.037M

Therefore the concentration after 3minutes is 1.037M

During an experiment, a student adds 2.90 g CaO to 400.0 mL of 1.500 M HCl . The student observes a temperature increase of 6.00 °C . Assuming that the solution's final volume is 400.0 mL , the density is 1.00 g/mL , and the heat capacity is 4.184 J/g⋅°C , calculate the heat of the reaction, ΔHrxn .

Answers

Answer:

ΔHrxn = 193107.69 J/mol

Explanation:

ΔHrxn = mcΔT

m = mass

c = heat capacity

ΔT = temperature variation

density = m/V

m = density x V

m = 1.00 g/mL x 400.0 mL

m = 400.0 g

ΔHrxn = mcΔT

ΔHrxn = 400 g x 4.184 J/g°C x 6.00 °C

ΔHrxn = 10041.6 J

CaO + 2HCl → CaCl₂ + H₂O

CaO = 56.0774 g/mol

2.90 g CaO = 0.052 mol

400.0 mL of 1.500 mol/L HCl = 0.6 mol HCl

ΔHrxn = 10041.6 J is for 0.052 mol of CaO

ΔHrxn = 193107.69 J is for 1 mol of CaO

Draw the Lewis structure for XeCl2 and answer the following questions.How many valence electrons are present in this compound?
How many bonding electrons are present in this compound?
How many lone pair (non-bonding) electrons are present in this compound?

Answers

Answer:

Valence electrons in XeCl2 = 8 + 7 + 7 = 22.

Bonding electrons = 4.

Nonbonding electrons = 18.

Explanation:

Hello.

In this case, you can see the Lewis structure on the attached picture, in which you can see that there are since xenon has 8 valance electrons and each chlorine has 7 valence electrons, the total amount of valence electrons is:

Valence electrons in XeCl2 = 8 + 7 + 7 = 22.

Moreover, since each chlorine atom is bonding with one of the eight electrons of xenon (Lewis structure), we can see there are 4 bonding electrons.

Finally, since there are six nonbonding electrons per chlorine atom and six nonbonding electrons in xenon, the overall nonbonding electrons are:

Nonbonding electrons in XeCl2 = 6 + 6 + 6 = 18.

Regards.

Read the descriptions below of two substances and an experiment on each. Decide whether the result of the experiment tells you the substance is a pure substance or a mixture, if you can. • Sample A is a solid yellow cube with a total mass of 50.0 g. The cube is put into a beaker filled with 250. mL of water. The cube collapses into a small pile of orange powder at the bottom of the beaker. When this powder is filtered out, dried and weighed, it has a total mass of 29.9 g. If the experiment is repeated with 500. mL of water, the powder that's left over has a mass of 10.0 g. Sample B is 100. g of a coarse grey powder with a faint unpleasant smell. 15. mg of the powder are put into a very thin tube and heated. The powder begins melting at 66.2 °C.The temperature stays constant as the powder slowly melts. After the last of the powder melts, the temperature starts to rise again, eventually reaching 76.0 °C. O pure substance Is sample A made from a pure substance or a mixture? x ? mixture If the description of the substance and the outcome of the experiment isn't enough to decide, choose "can't decide. (can't decide) O pure substance Is sample B made from a pure substance or a mixture? If the description of the substance and the outcome of bstance and the outcome of the experiment isn't enough to decide, choose "can't decide." mixture (can't decide)

Answers

Answer and Explanation: Sample A is a mixture. Solubility is characteristics of each substance, which means a substance can be distinguished from other substances and can be useful to separate mixtures.

In Sample A, when is added different volumes of water, the resulting powder has different mass. This means there are more than one substance forming the yellow cube. Therefore, sample A is a mixture.

Sample B is a puresubstance. Each substance has its own melting point. Whe na pure substance reaches its melting point, temperature is constant until all of that substance is melted. In sample B, temperature is stable at 66.2°C and then, after all the powder is melted, it rises again. Therefore, sample B is a pure substance.

Final answer:

Sample A is a mixture based on the experiment result, while the nature of Sample B is inconclusive.

Explanation:

The result of the experiment with sample A indicates that it is a mixture. When the solid yellow cube is put into water, it collapses into a small pile of orange powder. The mass of the powder that is left over depends on the amount of water used. This suggests that the cube is composed of different substances that can be separated by filtration.

On the other hand, the result of the experiment with sample B is inconclusive, so we can't decide whether it is a pure substance or a mixture. Heating the coarse grey powder causes it to melt at a constant temperature, but there is a temperature increase after the last of the powder melts. This could indicate that the powder is a pure substance with a melting point range, or it could suggest the presence of impurities.

Learn more about Substances and experiments here:

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Which compound is classified as an electrolyte? a) C6H12O6
b)Ca(OH)2
c)C12H22O11
d)CH3OH

Answers

Answer:
Option-B, Ca(OH)₂ is the correct answer.

Explanation:
By definition Electrolytes are those compounds which when poured in polar solvents dissociate into a positively charged specie called Cation and a negatively charged specie called an Anion. These ions have the potential to conduct electricity.
So Ca(OH)₂ is dissolved in water it produces Ca⁺² cation and OH⁻ anions. So it acts as an Electrolyte.
While, C₆H₁₂O₆ (Glucose), C₁₂H₂₂O₁₁ (Table Sugar) and CH₃OH (Methanol) are covalent compounds and does not produce any positive or negative ions when dissolved in water.

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