Bonus question #2: Unknown element, Element X, exists in nature as two isotopes:

If the rate law for a reaction A → P is rate = 3.37x10-3 M^-1 min-1 [A]^2 and the initial concentration of a is 0.122 M, calcula

DedPeter [7]

Rate = 3.37x10-3 M^-1 min-1 [A]^2 and the initial concentration of a is 0.122M.

A rate law indicates the rate of a chemical response depends on reactant concentration. For a response inclusive of the price regulation commonly has the form rate = ok[A]ⁿ, in which okay is a proportionality constant known as the fee regular and n is the order.

The charge of a chemical response is, perhaps, its maximum crucial asset because it dictates whether or not a reaction can arise all throughout an entire life. knowing the charge regulation, an expression concerning the price to the concentrations of reactants can assist a chemist to modify the response conditions to get an extra suitable rate.

half-life is the time taken for the radioactivity of a substance to fall to 1/2 its authentic cost whereas implies existence is the common life of all the nuclei of a particular risky atomic species.

Learn more about rate law here:-brainly.com/question/7694417

#SPJ4

3 0

2 years ago

How many moles would be equal to 2.6 liters of calcium nitrite

lorasvet [3.4K]

6 moles to equal that I do t know for sure though

7 0

3 years ago

A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C･g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('

polet [3.4K]

Answer:

$a=5.65cm$

Explanation:

Hello,

In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

$Q_{Pt}=-Q_{Deu}$

We can represent the heats in terms of mass, heat capacities and temperatures:

$m_{Pt}Cp_{Pt}(T_f-T_{Pt})=-m_{Deu}Cp_{Deu}(T_f-T_{Deu})$

Thus, we solve for the mass of platinum:

$m_{Pt}=\frac{-m_{Deu}Cp_{Deu}(T_f-T_{Deu})}{Cp_{Pt}(T_f-T_{Pt})} \\\\m_{Pt}=\frac{-1.00L*1110g/L*4.211J/(g\°C)*(41.9-25.5)\°C}{0.1256J/(g\°C)*(41.9-200.0)\°C} \\\\m_{Pt}=3860.4g$

Next, by using the density of platinum we compute the volume:

$V_{Pt}=\frac{3860.4g}{21.45g/cm^3}\\ \\V_{Pt}=180cm^3$

Which computed in terms of the edge length is:

$V=a^3$

Therefore, the edge length turns out:

$a=\sqrt[3]{180cm^3}\\ \\a=5.65cm$

Best regards.

6 0

3 years ago

A sample that contains only SrCO3 and BaCO3 weighs 0.846 g. When it is dissolved in excess acid, 0.234 g carbon dioxide is liber

In-s [12.5K]

Answer:28.605

Explanation:First, the molar mass of of SrCO3, BaCO3 and CO2 has to be calculated, (using the molar mass of each element Sr = 87.62, Ba = 137.327, C=12.011, O= 16.00)

The molar masses are;

SrCO3 = 87.62 + 12.011 + (3*16) = 147.631g/mol

BaCO3 = 79.904 + 12.011 + (3*16) = 197.34 g/mol

CO2 = 12.011 + (2*16) = 44.011 g/mol

To obtain one of the equations to solve the problem;

The sample is made of SrCO3 and BaCO3 and has a mass of 0.846 g. Representing the mass of SrCO3 as ma and that of BaCO3 as mb. The first equation can be written as:

ma + mb = 0.846g (1)

To obtain another equation in order to be able to determine the different percentages of the compounds (SrCO3 and BaCO3) that make of the sample, a relationship can be obtained by determining the relationship between the number of moles of CO2 formed as the mass of the SrCO3 and BaCO3;

The number of moles of CO2 formed = (mass of CO2)/(molar mass) =0.234/44.011 =0.00532moles

CO2 contains 1 mole of carbon (C) so therefore 0.00532 moles of CO2 contains 0.00532 moles of C

The sample produced 0.00532 moles of CO2, therefore the number of moles SrCO3 and BaCO3 that produced this amount can be calculated using the formula;

= (mass )/(molar mass)

No of moles of SrCO3 and BaCO3 will be ma/147.631 and mb/197.34 moles respectively

The total amount of C molecules produced by SrCO3 and BaCO3 will be 0.00532 moles of C

The second equation can be written as

ma/147.631 + mb/197.34= 0.00532 (2)

Solving Equation (1) and (2) simultaneously;

ma = 0.604g; mb = 0.242g

Therefore the percentage of BaCO3 = (mass of BaCO3 )/(mass of sample )*100

= 0.242/(0.846 )*100

= 28.605%

5 0

3 years ago

Question 1 (2 points)

bogdanovich [222]

Answer:

3.75 L

Explanation:

We can solve this problem by using Charles' law, which states:

V₁T₂=V₂T₁

Where subscript 1 stands for initial volume and temperature and subscript 2 for final volume and temperature, meaning that in this case:

V₁ = 2.5 L

T₂ = 300 K

V₂ = ?

T₁ = 200 K

We input the data:

2.5 L * 300 K = V₂ * 200 K

And solve for V₂:

V₂ = 3.75 L

5 0

3 years ago