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Environmental pollution with chemicals <br> how does it affect countries and economies

netineya [11]

Answer:

The most recent Global Burden of

Disease (GBD) study estimates that air pollution – indoor and outdoor combined – was the cause of 5.5 million premature deaths

globally in 2013. Air pollution also has further consequences on human health, leading in particular to an increasing number of

respiratory and cardiovascular diseases. Moreover, it affects crop yields and the environment, with impacts on biodiversity and

ecosystems, amongst others. These impacts have significant economic consequences, which will affect economic growth as well as welfare

Explanation:

5 0

2 years ago

Select the conjugate acid-base pair(s). a) HI, I b) HCHO2, SO4^2- c) CO3^2-, HCI d) PO4^3-, HPO4^2-

Semenov [28]

Answer:

d) PO4^3-, HPO4^2-

Explanation:

Basically, an acid and a base which differs only by the presence or absence of proton (hydrogen ion) are called a conjugate acid-base pair.

a) HI, I

This is incorrect. For the acid, HI the conjugate base is I⁻ ion.

b) HCHO2, SO4^2-

This is incorrect, there's no relationship between both entities.

c) CO3^2-, HCI

This is incorrect, there's no relationship between both entities.

d) PO4^3-, HPO4^2-

This is correct. The difference between both entities is the Hydrogen ion. This is the conjugate acid-base pair

6 0

3 years ago

Current Question:

IRISSAK [1]

Answer:

0.2g

Explanation:

All radiodecay follows the 1st order decay equation

A = A₀e^-kt

A => Activity at time (t)

A₀ => Initial Activity at time = 0

k => decay constant for isotope

T => time in units that match the decay constant

Half-Life Equation => kt(½) = 0.693 => k = 0.693/34 min = 0.0204min¹

A = A₀e^-kt = (26g)e^-(0.0204/min)(238min) = (26g)(0.0078) = 0.203g ~ 0.2g (1 sig fig).

7 0

3 years ago

Calculate the heat change involved when 2.00 L of water is heated from 20.0/C to 99.7/C in

Pani-rosa [81]

Answer:

666,480 Joules or 669.48 kJ

Explanation:

We are given;

Volume of water as 2.0L or 2000 ml

but, density of water is 1 g/ml

Therefore, mass of water is 2000 g

Initial temperature as 20 °C
Final temperature as 99.7° C

Required to determine the heat change

We know that ;

Heat change = Mass × Temperature change × specific heat

In this case;

Specific heat of water is 4.2 J/g°C

Temperature change is 79.7 °C

Therefore;

Heat change = 2000 g × 79.7 °C × 4.2 J/g°C

= 669,480 Joules 0r 669.48 kJ

Thus, the heat change involved is 666,480 Joules or 669.48 kJ

5 0

3 years ago

At 298 K, what is the Gibbs free energy change (ΔG) for the following reaction?

9966 [12]

Answer:

(a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

Explanation:

Given that,

Temperature = 298 K

Suppose, density of graphite is 2.25 g/cm³ and density of diamond is 3.51 g/cm³.

$\Delta H\ for\ diamond = 1.897 kJ/mol$

$\Delta H\ for\ graphite = 0 kJ/mol$

$\Delta S\ for\ diamond = 2.38 J/(K mol)$

$\Delta S\ for\ graphite = 5.73 J/(K mol)$

(a) We need to calculate the value of $\Delta G$ for diamond

Using formula of Gibbs free energy change

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G= (1897-0)-298\times(2.38-5.73)$

$\Delta G=2895.3$

$\Delta G=2.895\ kJ$

The Gibbs free energy change is positive.

(b). When it is compressed isothermally from 1 atm to 1000 atm

We need to calculate the change of Gibbs free energy of diamond

Using formula of gibbs free energy

$\Delta S=V\times\Delta P$

$\Delta S=\dfrac{m}{\rho}\times\Delta P$

Put the value into the formula

$\Delta S=\dfrac{12\times10^{-6}}{3.51}\times999\times10130$

$\Delta S=34.59\ J/mole$

(c). Assuming that graphite and diamond are incompressible

We need to calculate the pressure

Using formula of Gibbs free energy

$\beta= \Delta G_{g}+\Delta G+\Delta G_{d}$

$\beta=V(-\Delta P_{g})+\Delta G+V\Delta P_{d}$

$\beta=\Delta P(V_{d}-V_{g})+\Delta G$

Put the value into the formula

$0=\Delta P(\dfrac{12\times10^{-6}}{3.51}-\dfrac{12\times10^{-6}}{2.25})\times10130+2895.3$

$0=-0.0194\Delta P+2895.3$

$\Delta P=\dfrac{2895.3}{0.0194}$

$\Delta P=14924\ atm$

(d). Here, $C_{p}=0$

So, The value of $\Delta H$ and $\Delta S$ at 900 k will be equal at 298 K

We need to calculate the Gibbs free energy of diamond relative to graphite

Using formula of Gibbs free energy

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G=(1897-0)-900\times(2.38-5.73)$

$\Delta G=4912\ J$

Hence, (a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

7 0

3 years ago