Ga3+ and Br1- is what formula?

explain the law of conservative of energy, give a specific example using kinetic and potential energy that shows how energy is c

KiRa [710]

Answer:

The Law of Conservation of Energy states that energy cannot be created or destroyed. In other words, the total energy of a system remains constant. This is an important concept to remember when dealing with energy problems. The two basic forms of energy that we will focus on are kinetic energy and potential energy.

Explanation:

In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. This law means that energy can neither be created nor destroyed; rather, it can only be transformed or transferred from one form to another.

Im bad at these questions hope it helps and have a good day.

8 0

3 years ago

How many moles are in 20g of potassium (K)?

otez555 [7]

Explanation:

Hi for this one u just need to remember and use the equation.

$moles = \frac{mass}{mr}$

then u find mr of potassium which is 39.1.

then u do

$\frac{20}{39.1}$

you get the answer as 0.5115 write ur answer to 3 significant figures which will be 0.512 moles .

hope this helps :)

7 0

3 years ago

Transuranium elements are artificial and may be prepared by a fusion process in which heavy nuclei are bombarded with light ones

Scilla [17]

Answer:

$_{97}^{345}\textrm{Bk}$

Explanation:

In a nuclear reaction, the total mass and total atomic number remains the same.

Am has an atomic number of 95. So correct reaction is:-

$^{343}_{95}\textrm{Am}+^{4}_{2}\textrm{He}\rightarrow ^A_Z\textrm{X}+2^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

343 + 4 = A + 2

A = 345

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

95 + 2 = Z + 0

Z = 97

Hence, the isotopic symbol of unknown element is $_{97}^{345}\textrm{Bk}$

7 0

3 years ago

or a particular isomer of C 8 H 18 , the combustion reaction produces 5104.1 kJ of heat per mole of C 8 H 18 ( g ) consumed, und

beks73 [17]

Answer:

The standard enthalpy of formation of this isomer of $C_{8}H_{18}$ is -220.1 kJ/mol.

Explanation:

The given chemical reaction is as follows.

$C_{8}H_{18}(g)+ \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g)+9H_{2}O(g)$

$\Delta H^{o}_{rxn}= -5104.1kJ/mol$

The expression for the entropy change for the reaction is as follows.

$\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]$

$\Delta H^{o}_{f}(H_{2}O)= -241.8kJ/mol$

$\Delta H^{o}_{f}(CO_{2})= -393.5kJ/mol$

$\Delta H^{o}_{f}(O_{2})= 0kJ/mol$

Substitute the all values in the entropy change expression.

$-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]$

$-5104.1kJ/mol=-5324.2kJ/mol -\Delta H^{o}_{f}(C_{8}H_{18})$

$\Delta H^{o}_{f}(C_{8}H_{18}) =-5324.2kJ/mol +5104.1kJ/mol$

$=-220.1kJ/mol$

Therefore, The standard enthalpy of formation of this isomer of $C_{8}H_{18}$ is -220.1 kJ/mol.

4 0

3 years ago

A closed vessel system of volume 2.5 L contains a mixture of neon and fluorine. The total pressure is 3.32 atm at 0.0°C. When th

gayaneshka [121]

Answer:

moles Ne = 0.154 mol

moles F₂ = 0.217 mol

Explanation:

Step 1: Data given

Volume of the vessel system = 2.5 L

Total pressure = 3.32 atm at 0.0 °C

The mixture is heated to 15.0 °C

The entropy of the mixture increases by 0.345 J/K

The heat capacity of monoatomic gas = 3/2R and that for a diatomic gas = 5/2R

Step 2: Define the gas

Neon is a monoatomic gas, composed of Ne atoms

⇒ Cv(Ne) ≅ (3/2)R

Fluorine is a diatomic gas, composed of F₂ molecules.

⇒ Cv(F₂) ≅ (5/2)R

Step 3: Calculate moles of gas

p*V = n*R*T

⇒ with p = the total pressure = 3.32 atm

⇒ with V = the total volume = 2.5 L

⇒ with n = the number of moles of gas

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 273.15 Kelvin

n(total) = p*V/RT = (3.32 atm*2.5 L)/(0.08206 L*atm/mol•K*273.15) = 0.3703 mol

Step 4: Calculate moles of Ne and F2

For one mole heated at constant volume,

∆S = Cv*ln(288.15/273.15) = 0.05346*Cv

⇒ ∆S for 0.3703 mol,

∆S = (0.3703 mol)(0.05346)Cv = 0.345 J/K

⇒ Cv = 17.43 J/mol*K for the Ne/F₂ mixture.

For pure Ne, Cv = (3/2)R = 1.5*8.314 J/mol*K = 12.471 J/mol*K

For pure F₂, Cv = (5/2)R = 2.5 * 8.314 J/mol*K = 20.785 J/mol*K

if X is the mole fraction of Ne, we can find X by:

17.43 J/mol*K = X* 12.471 J/mol*K + (1 – X) * 20.785 J/mol*K

⇒ 20.875 – 8.314 * X = 17.43

X = 0.415 , 1 – X = 0.585

moles Ne = (0.415)(0.3703 mol) = 0.154 mol

moles F₂ = (0.585)(0.3703 mol) = 0.217 mol

4 0

4 years ago