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3 years ago

If the toxic quantity is 1.5 gg of ethylene glycol per 1000 gg of body mass, what percentage of ethylene glycol is fatal

Chemistry

1 answer:

Elina [12.6K]3 years ago

4 0

Answer:

The percentage of ethylene glycol that is fatal is 0.15 %

Explanation:

If the toxic quantity of ethylene glycol in a 1kg or 1000 g body weight is 1.5 g then the percentage of ethylene glycol that is fatal is

$\frac{1.5}{1000}$ ˣ 100 = 0.15%

Hence, the percentage of ethylene glycol that is toxic for any body weight is 0.15%. This percentage is very important in various aspects of science including drug discovery and food production/processing

You might be interested in

How many polar bonds are found in this molecule

Law Incorporation [45]

A bond is non polar if it is between same atoms and polar if it is between different atoms.

Same atoms are like two dogs of same strength pulling a bone towards towards each other. But when it’s different atoms it’s like a big dog and small dog then the bone is more towards bigger dog. So it’s the same way in bonds.

Bonds are made up of electrons, when the more stronger pulling atom is present than other the electrons are more towards it and as a result we have polar bond. There is development of a kind of a negative pole and a positive pole.

The stronger atom has electrons towards itself so it has a little more negative charge while the other atom has positive charge. This makes bond polar.

So just look for bond between two different atoms, it would be polar.

Look at the pic below to see the answer.

Marked with green is bond between same atoms... one carbon and another carbon so it is not polar and test marked with blue are polar.

Well the answer should have been 10 but since the bonds at 3 and 8 are two of same type we count only one of them.

The answer is 8... well the answer should be 10 otherwise... discuss it with ur teacher

6 0

3 years ago

calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w

kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is $-22Jmole^-^1k^-^1$

(b) The normal boiling point of water (J·K−1·mol−1) is $-109Jmole^-^1K^-^1$

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is 109J/mole

Explanation:

Lets calculate

(a) - General equation -

$(\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p$ = $-5_m(\beta )+5_m(\alpha )$ = $-\frac{\Delta H}{T}$

$\alpha ,\beta$ → phases

ΔH → enthalpy of transition

T → temperature transition

$(\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p$ = $= -\frac{\Delta_fH}{T_f}$

= $\frac{-6.008kJ/mole}{273.15K}$ ( $\Delta_fH$ is the enthalpy of fusion of water)

= $-22Jmole^-^1k^-^1$

(b) $(\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}$

= $\frac{40.656kJ/mole}{373.15K}$ ( $\Delta_v_a_p_o_u_rH$ is the enthalpy of vaporization)

= $-109Jmole^-^1K^-^1$

$[\mu(l-5$ ° $C)-\mu(l,0$ ° $C)]$ = $[\mu(s-5$ ° $C)-\mu(s,0$ ° $C)]$ $=-S_m$ ΔT

$\mu(l,-5$ ° $C)-\mu(s,-5$ ° $C)=-Sm\DeltaT [\mu(l,0$

$\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)$

= 109J/mole

6 0

2 years ago

If you have 100 ml of a 0.10 m tris buffer (pka 8.3) at ph 8.3 and you add 3.0 ml of 1.0 m hcl, what will be the new ph?

sukhopar [10]

The new pH is 7.69.

According to Hendersen Hasselbach equation;

The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution and the pKa or pKb and the ratio of the concentrations of the dissociated chemical species. To calculate the pH of the buffer solution made by mixing salt and weak acid/base. It is used to calculate the pKa value. Prepare buffer solution of needed pH.

pH = pKa + log10 ([A–]/[HA])

Here, 100 mL of 0.10 m TRIS buffer pH 8.3

pka = 8.3

0.005 mol of TRIS.

∴ $8.3 = 8.3 + log \frac{[0.005]}{[0.005]}$

<em> </em>inverse log 0 = $\frac{[B]}{[A]}$

$\frac{[B]}{[A]} = 1$

Given; 3.0 ml of 1.0 m hcl.

pka = 8.3

0.003 mol of HCL.

$pH = 8.3 + log \frac{[0.005-0.003]}{[0.005+0.003]}\\pH = 8.3 + log \frac{[0.002]}{[0.008]}\\\\pH = 8.3 + log {0.25}\\\\pH = 8.3 + (-0.62)\\pH = 7.69$

Therefore, the new pH is 7.69.

Learn more about pH here:

brainly.com/question/24595796

#SPJ1

8 0

2 years ago

What causes water to become denser when it is carried to the poles by surface currents? increased temperature and increased sali

Anarel [89]

Answer:

The answer is decreased temperature and increased salinity

Explanation:

It is what is known as the thermohaline circulation

The thermohaline circulation moves the water slowly. This water moves mainly due to differences in its relative density. Much denser water sinks over water that is less dense. Two factors impact the density of seawater: temperature and salinity.

Cold water is denser than hot water:

-Water cools when it loses heat, it occurs at high latitudes.

-Water is heated when it receives energy from the sun, at low latitudes.

Saltier water is much denser than water that has less salt:

-Sea water becomes salty if the evaporation rate increases.

-Sea water becomes less salty if there is a water inlet over the sea.

3 0

3 years ago

Read 2 more answers

An archer pulls her bowstring back 0.370 m by exerting a force that increases uniformly from zero to 264 N. (a) What is the equi

Arte-miy333 [17]

Answer:

713.51 N/m

Explanation:

Hook's Law: This law states that provided the elastic limit is not exceeded, the extension in an elastic material is directly proportional to the applied force.

From hook's law,

F = ke ...........................Equation 1

Where F = Force exerted on the bowstring, e = Extension/compression of the bowstring, k = Spring constant of the bow.

Make k the subject of the equation,

k = F/e ............................ Equation 2

Given: F = 264 N, e = 0.37 m.

Substitute into equation 2

k = 264/0.37

k = 713.51 N/m

Hence the spring constant of the bow = 713.51 N/m

3 0

3 years ago