What is the double bond of XeF4

When an atom undergoes beta decay, wich of the following is true?

tigry1 [53]

The atomic number increases by one and the element becomes a different element.

3 0

3 years ago

What is the emperical formula of a compound containing 40.1% carbon and 6.6% hydrogen

levacccp [35]

Answer:

CH₂

Explanation:

Given parameters:

Percentage composition:

Carbon = 40.1%

Hydrogen = 6.6%

Unknown:

Empirical formula of the compound = ?

Solution:

The empirical formula of a substance is its simplest formula.

Elements Carbon Hydrogen

Percentage

Composition 40.1 6.6

Molecular mass 12 1

Number of moles 40.1/12 6.6/1

3.342 6.6

Divide through by

the smallest 3.342/3.342 6.6/3.342

1 2

So the empirical formula of the compound is CH₂

5 0

2 years ago

A rigid vessel contains 3.98 kg of refrigerant-134a at 700 kPa and 60°C. Determine the volume of the vessel and the total intern

Arada [10]

Answer: The volume of the vessel is $0.1542m^3$ and total internal energy is 162.0 kJ.

Explanation:

To calculate the volume of water, we use the equation given by ideal gas, which is:

$PV=nRT$

or,

$PV=\frac{m}{M}RT$

where,

P = pressure of container = 700 kPa

V = volume of container = ? L

m = Given mass of R-134a = 3.98 kg = 3980 g (Conversion factor: 1kg = 1000 g)

M = Molar mass of R-134a = 102.03 g/mol

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

T = temperature of container = $60^oC=[60+273]K=333K$

Putting values in above equation, we get:

$700kPa\times V=\frac{3980g}{102.03g/mol}\times 8.31\text{L kPa }\times 333K\\\\V=154.21L$

Converting this value into $m^3$ , we use the conversion factor:

$1m^3=1000L$

So, $\Rightarrow (\frac{1m^3}{1000L})\times 154.21L$

$\Rightarrow 0.1542m^3$

To calculate the internal energy, we use the equation:

$U=\frac{3}{2}nRT$

or,

$U=\frac{3}{2}\frac{m}{M}RT$

where,

U = total internal energy

m = given mass of R-134a = 3.98 kg = 3980 g (Conversion factor: 1kg = 1000g)

M = molar mass of R-134a = 102.03 g/mol

R = Gas constant = $8.314J/K.mol$

T = temperature = $60^oC=[60+273]K=333K$

Putting values in above equation, we get:

$U=\frac{3}{2}\times \frac{3980g}{102.03g/mol}\times 8.314J/K.mol\times 333K\\\\U=161994.6J$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, 161994.6 J = 162.0 kJ

Hence, the volume of the vessel is $0.1542m^3$ and total internal energy is 162.0 kJ.

6 0

3 years ago

Name 4 invasive species and describe their effect on ecosystem.

Grace [21]

Spotted Lantern: Destroying our trees

Fishers: Brought to lower numbers of certain species, but have dramatically lowered other species.

Zebra Mussels: Clog pipes

Emerald Ash Borer: Ruins ash trees that provide us with durable wood.

6 0

3 years ago

The pka of hco3 - (or pka2 of h2co3) is 10.33. what is the ph of a solution that has 0.1 m na2co3 and 1.0 m nahco3?

Margarita [4]

The given solution is a mixture of $NaHCO_{3}$ and $Na_{2}CO_{3}$ . It acts as a buffer as it is a combination of the weak acid $HCO_{3}^{-}$ and it's conjugate base $CO_{3}^{2-}$ .

$[Acid] = [HCO_{3}^{-}]=1.0 M$

$[Base] =[CO_{3}^{2-}]=0.1 M$

pH of the buffer solution can be calculated from Hendersen-Hasselbalch equation as below:

$pH=pK_{a}+log\frac{[CO_{3}^{2-}]}{[HCO_{3}^{-}]}$

$pH=10.33+log\frac{0.1}{1.0}$

pH = 10.33+(-1)

= 9.33

4 0

3 years ago