Answer:
Option C
CH₃CH₂CH₂COOH
Explanation:
Carbonxylic acids are compounds which has the general formula
R–COOH where R is an alkyl group.
Considering the options given in the question above,
For A:
CH₃CH₂OCH₂CH₃ is an ether compound with general formula ROR' where R and R' are both alkyl group.
For B:
CH₃CH₂CH₂CH₂OH is an alcohol with general formula ROH where R is an alkyl group.
For C:
CH₃CH₂CH₂COOH is a carbonxylic acid with general formula R–COOH where R is an alkyl group.
For D:
CH₃CH₂C=OCH₂CH₃ is a ketone compound with general formula RC=OR' where R and R' are both alkyl group
For E:
ClCH₂CH₂CH₂CH₂CH₂CH₂Br is simply an Alkyl halide with general formula XRX where X is an halogen (i.e F, Cl, Br or I) and R is an alkyl group.
From the above illustration, only option C contains a Carbonxylic compound.
Answer:
lattice parameter = 5.3355x10^-8 cm
atomic radius = 2.3103x10^-8 cm
Explanation:
known data:
p=0.855 g/cm^3
atomic mass = 39.09 g/mol
atoms/cell = 2 atoms
Avogadro number = 6.02x10^23 atom/mol
a) the lattice parameter:
Since potassium has a cubic structure, its volume is equal to:
v = [(atoms/cell)x(atomic mass)/(p)x(Avogadro number)]
substituting values:
v =[(2)x(39.09)/(0.855x6.02x10^23)]=1.5189x10^-22 cm^3
but as the cell volume is
a^3 =v
cm
for a BCC structure, the atomic radius is equal to

Answer:

Explanation:
Any gas at standard temperature and pressure (STP) has a volume of 22.4 liters per mole or 22.4 L/mol. We can create a proportion with this value.

Multiply both sides of the equation by 6.8 moles of krypton.

The units of moles of krypton will cancel.

The denominator of 1 can be ignored, so this becomes a simple multiplication problem.


If we round to the nearest whole number, the 3 in the tenths place tells us to leave the 2 in the ones place.

6.8 moles of krypton gas at standard temperature and pressure is equal to <u>152 liters</u>.
Answer:
354.67K
Explanation:
Applying
P₁V₁/T₁ = P₂V₂/T₂................. Equation 1
Where Where P₁ = initial pressure, T₁ = Initial temperature, V₁ = Initial Volume, P₂ = Final pressure, V₂ = Final Volume, T₂ = Final Temperature.
From the question, we are ask to look for the final temperature,
Therefore we make T₂ the subject of the equation
T₂ = P₂V₂T₁/P₁V₁............. Equation 2
Given: P₁ = 600 kPa, V₁ = 500 mL, T₁ = 77 °C = (273+77) = 350 K, P₂ = 760 kPa, V₂ = 400.0 mL
Substitute these values into equation 2
T₂ = (760×400×350)/(600×500)
T₂ = 354.67 K
Answer:
i assume that it would be a gametophyte.
Explanation: