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2 years ago

Question 2 (5 points)

Chemistry

1 answer:

blagie [28]2 years ago

5 0

The water in Glass A is cooler than the water in Glass B; therefore, the particles in Glass A move slower.

Option D

<h3>Chemical Reactions</h3>

Generally,the experiment shows that glass B temperature is higher than glass temperature A and this is given that observation that the solute dissolves faster in glass B than glass A.

Therefore,The water in Glass A is cooler than the water in Glass B; therefore, the particles in Glass A move slower.

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A gas cylinder is filled with 4.00 moles of oxygen gas at 300.0 k. the piston is compressed to yield a pressure of 400.0 kpa. wh

laila [671]

We can use the ideal gas law equation to find the volume of the gas.
PV = nRT
P - pressure - 400 kPa
V - volume
n - number of moles - 4.00 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 300.0 K
substituting these values in the equation
400 000 Pa x V = 4.00 mol x 8.314 Jmol⁻¹K⁻¹ x 300.0 K
V = 24.9 dm³
Volume is 24.9 dm³

5 0

3 years ago

Must new substances be formed when you observe a chemical property explain

GaryK [48]

No but a chemical change yes a chemical property is only a characteristic not change

5 0

3 years ago

Lead(II) nitrate is added slowly to a solution that is 0.0800 M in Cl− ions. Calculate the concentration of Pb2+ ions (in mol /

barxatty [35]

Answer:

$[Pb^{2+}]=3.9 \times 10^{-2}M$

this is the concentration required to initiate precipitation

Explanation:

$PbCl_2$ ⇄ $Pb^{2+}+2Cl^-$

Precipitation starts when ionic product is greater than solubility product.

Ip>Ksp

Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.

This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.

$Ip=[Pb^{2}][2Cl^-]^2=Ksp$

$Ksp=2.4\times 10^{-4}$

lets solubility=S

$[Pb^{2+}] = S$

$[Cl^-]=2S$

$Ksp=[Pb^{2+}]\times [Cl^-]^2$

$Ksp=S \times (2S)^2$

$Ksp=4S^3$

$S=\sqrt[3]{\frac{Ksp}{4} }$

$S=3.9\times 10^{-2}$

$[Pb^{2+}]=3.9 \times 10^{-2}M$ this is the concentration required to initiate precipitation

4 0

3 years ago

What are the empirical and molecular formulas of a hydrocarbon if combustion of 2.10 g of the compound yields 6.59 g co2 and 2.7

mariarad [96]

The empirical formula of hydrocarbon is CH2

The molecular formula of the hydrocarbon is C6H12

<u><em>Explanation</em></u>

Hydrocarbon is made up of carbon and hydrogen

<h3><u><em> </em></u>Empirical formula calculation</h3>

Step 1: find the moles CO2 and H2O

moles =mass/molar mass

moles of CO2 = 6.59 g/ 44 g/mol = 0.15 moles

moles of H2O = 2.70 g / 18 g/mol = 0.15 moles

Step 2: Find the moles ratio of Co2:H2O by diving each mole by smallest mole(0.15)

that is for CO2 = 0.15/0.15 =1

For H2O = 0.15/0.15 =1

therefore the mole ratio of Co2 : H2O = 1:1 which implies that 1 mole of Co2 and 1 mole of H2O is formed during combustion reaction.

From the the law of mass conservation the number of atoms in reactant side must be equal to number of atoms in product side

therefore since there 1 atom of C in product side there must be 1 atom of C in reactant side.

In addition there is 2 H atom in product side which should be the same in reactant side.

From information above the empirical formula is therefore = CH2

Molecular formula calculation

[CH2}n= 84 g/mol

[12+ (1x2)] n = 84 g/mol

14 n = 84 g/mol

n = 6

multiply the each subscript in CH2 by 6

Therefore the molecular formula = C6H12

5 0

3 years ago

For a steel stack that exhausts 1,200 m3/min of gases at 1 atm and 400 k, calculate the inner diameter if you are designing for

bonufazy [111]

The inner diameter for a steel stack that exhausts 1,200 m3/min of gases at 1 atm and 400 k is 1.45 m

<h3>What is Stack Height ?</h3>

Stack height means the distance from the ground-level elevation at the base of the stack to the crown of the stack.

If a stack arises from a building or other structure, the ground-level elevation of that building or structure will be used as the base elevation of the stack.

Given is a steel stack that exhausts 1,200 cu.m/min of gases

P= 1 atm and

T= 400 K

maximum expected wind speed at stack height of 12 m/s

The formula for the diameter of chimney

$\rm d=\sqrt{\dfrac{4Q}{\pi v} }$

Q =1200 cu.m/min

= 1200 * 0.0166 = 19.92 cu.m/sec

Velocity = 12m/s

$\rm d=\sqrt{\dfrac{4\times 19.92}{3.14*12} }\\$

d= 1.45 m

Therefore The inner diameter for a steel stack that exhausts 1,200 m3/min of gases at 1 atm and 400 k is 1.45 m.

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8 0

2 years ago