Are babies producers or consumers or none? What about rocks? Are they producers or consumers?

Iodine-131 decays with a half-life of 8.02

dlinn [17]

Radioactive material undergoes 1st order decay kinetics.

For 1st order decay, half life = 0.693/k

where k = rate constant

k = 0.693/half life = 0.693/8.02 = 0.0864 day-1

Now, for 1st order reaction,
k = $\frac{2.303}{t} X log \frac{initial.conc}{final.conc}$

Given: t = 6.01d, initial conc. = 5mg

∴0.0864 = $\frac{2.303}{6.01} X log \frac{5}{final.conc}$
∴ final conc. = 2.975 mg

3 0

3 years ago

(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct

Zarrin [17]

Explanation:

The given data is as follows.

$\Lambda^{o}_{m}$ (NaCl) = $1.264 \times 10^{-2}$

$\Lambda^{o}_{m}$ (H-O=C-ONO) = $1.046 \times 10^{-2}$

$\Lambda^{o}_{m}$ (HCl) = $4.261 \times 10^{-2}$

Conductivity of monobasic acid is $5.07 \times 10^{-2} S m^{-1}$

Concentration = 0.01 $mol/dm^{3}$

Therefore, molar conductivity ( $\Lambda_{m}$ ) of monobasic acid is calculated as follows.

$\Lambda_{m} = \frac{conductivity}{concentration}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}$

= $5.07 \times 10^{-3} S m^{2} mol^{-1}$

Also, $\Lambda^{o}_{m}$ = $\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}$

= $4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}$

= $4.043 \times 10^{-2} S m^{2} mol^{-1}$

Relation between degree of dissociation and molar conductivity is as follows.

$\alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}$

= 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

Putting the values into the above formula we get the following.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

= $\frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}$

= $0.017973 \times 10^{-2}$

= $1.7973 \times 10^{-4}$

Hence, the acid dissociation constant is $1.7973 \times 10^{-4}$ .

Also, relation between $pK_{a}$ and $K_{a}$ is as follows.

$pK_{a} = -log K_{a}$

= $-log (1.7973 \times 10^{-4})$

= 3.7454

Therefore, value of $pK_{a}$ is 3.7454.

3 0

2 years ago

Draw the Lewis structure for the compound with the formula COCl2COCl2. Use lines to show bonding electrons.

stira [4]

Answer : The Lewis-dot structure of $COCl_2$ is shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, $COCl_2$

As we know that carbon has '4' valence electrons, chlorine has '7' valence electron and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in $COCl_2$ = 1(4) + 2(7) + 1(6) = 24

According to Lewis-dot structure, there are 8 number of bonding electrons and 16 number of non-bonding electrons.

6 0

3 years ago

I NEED HELP HERE PLEASE !!!

Juliette [100K]

Answer:

Gas it too expensive I need to get a job. I have no money to use at the atm

Explanation:

if your car is on 7000 ml it has issues and you need to take it to a mechanic.

4 0

2 years ago

The empirical formula of styrene is CH; find the % composition for H and CH. How many atoms are present in a 2.00 gram sample of

lina2011 [118]

% Composition = ?
no atom in 2 g of styrene =?
molar mass of strene =104.15
% composition of c= 12/13.008 =.922*100=92.2
% composition of h =1.008/13.008=0.0774*100= 7.74

no gram atom=mass in kg /molar mass=2/104.15 =0.01920 mol
no of gram atom * avogadro's number = 0.0192*6.02 *10( exponent 23) =1.15584

8 0

2 years ago