What is the capacity to do work known as?

Which of the following pairs of elements belong to the same period?

bagirrra123 [75]

Answer:

He and Ar

Explanation:

3 0

3 years ago

What is the specific heat of titanium in j/(g⋅∘c) if it takes 89.7 j to raise the temperature of a 33.0 g block by 5.20 ∘c?

larisa [96]

Specific heat capacity is the amount of energy required to raise one gram of substances by 1 degree celsius . Therefore specific heat capacity for tatanium is 89.7j /( 33.0g x5.2 degree celsius) = 0.52j/g degree celcius
Molar mass for tatanium is 47.9 g/mole
heat is therefore 47.9 g/mole x 0.52j/g =24.9j/mole

5 0

2 years ago

Read 2 more answers

Consider the reaction below. Initially the concentration of SO2Cl2 is 0.1000 M. Solve for the equilibrium concentration of SO2Cl

tensa zangetsu [6.8K]

Answer:

$[SO_2Cl_2] = 0.09983 M$

Explanation:

Write the balance chemical equation ,

$SO_2Cl_2((g) = SO_2(g) + Cl_2(g)$

initial concenration of $SO_2Cl_2((g) =0.1M$

lets assume that degree of dissociation= $\alpha$

concenration of each component at equilibrium:

$[SO_2Cl_2] = 0.1-0.1\alpha$

$[SO_2] = 0.1\alpha$

$[Cl_2] = 0.1\alpha$

$Kc =\frac{0.1\alpha \times 0.1\alpha}{0.1-0.1\alpha}$

$Kc =\frac{0.1\alpha \times \alpha}{1-\alpha}$

as $\alpha$ is very small then we can neglect $1-\alpha$

therefore ,

$Kc ={0.1\alpha \times \alpha}$

$\alpha =\sqrt{\frac{Kc}{0.1}}$

$\alpha = 1.73 \times 10^{-3}$

Eqilibrium concenration of $[SO_2Cl_2] = 0.1-0.1\alpha = 0.1-0.1\times 0.00173$

$[SO_2Cl_2] = 0.09983 M$

4 0

2 years ago

Acetylene burns in air according to the following equation: C2H2(g) + 5 2 O2(g) → 2 CO2(g) + H2O(g) ΔH o rxn = −1255.8 kJ Given

professor190 [17]

Answer: -227 kJ

Explanation:

The balanced chemical reaction is,

$C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]$

where,

n = number of moles

$\Delta H_{O_2}=0$ (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

$-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]$

$-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]$

$\Delta H_{C_2H_2}=-227kJ$

Therefore, the enthalpy change for $C_2H_2$ is -227 kJ.

7 0

2 years ago

Five gases combined in a gas cylinder have the following partial pressures: 3. 00 atm (N2), 1. 80 atm (O2), 0. 29 atm (Ar), 0. 1

Helen [10]

The total pressure of the gaseous mixture has been 5.37 atm. Thus, option D is correct.

The partial pressure has been defined as the pressure exerted by each gas in the mixture.

According to the Dalton's law of partial pressure, the total pressure of gas has been the sum of the partial pressure of the gases in the mixture.

The given partial pressure of gases in the mixture has been: