All categories

2 years ago

1 point

1 answer:

8 0

I'm pretty sure its D because if it's manufactured and distributed its likely not toxin or poison. a narcotic could be a controlled substance so that's why I'm leaning towards D

You might be interested in

Median of 4, 6 ,9 helppp

aniked [119]

For the data set 1, 1, 2, 5, 6, 6, 9 the median is 5. For the data set 1, 1, 2, 6, 6, 9 the median is 4.

hope it helps you

3 0

3 years ago

Read 2 more answers

Which one is more metallic? A) Gold or B) Bismuth

Arlecino [84]

Answer:

Gold

Explanation:

Download pdf

6 0

3 years ago

Carbon and silicon are tetravalent but ge sn pb show di valency. Why

Anuta_ua [19.1K]

Answer:

Carbon and silicon both are tetravalent elements as compared to germanium, tin, and lead which are divalent.

That's because Ge, tin, and Pb show inert pair effect and has a greater nuclear effective charge on the 's' electrons due to poor shielding effect. .That's why these elements are not able to share their valence electrons while carbon and silicon does and show "catenation" which is the ability to form long chain molecules.

4 0

3 years ago

One mole of an ideal gas with a volume of 1.0 L and a pressure of 5.0 atm is allowed to expand isothermally into an evacuated bu

Deffense [45]

Answer:

w= - 1.7173 kJ, q= 1.7173 kJ, q(rev) = 1717.3 J = 1.7173 kJ.

Explanation:

Okay, from the question we are given the information below;

Number of moles, n= 1 mole; initial volume, v(1) = 1.0 litres (L); pressure (p) = 5atm, final volume(v2) = 2.0 Litres(L) ; the workdone, w= not given; the heat, q and q(rev)= not given and the gas was said to expand isothermally.

So, this question is a question from the part of chemistry known as thermodynamics. Therefore, grip yourself we are delving into thermodynamics 'waters' now.

For expansion isothermally; the workdone, w= -nRT ln v2/v1.

Where T= temperature= 25° C = 298 k and R= gas constant.

Therefore; workdone, w = - 1 × 8.314 × 298 × ln(2/1).

Workdone,w= - 1717.32204643. =

- 1717.3 Joules (J).

==> Workdone,w= - 1.7173 kJ.

Then, we are to find q. q can be solved by using the first law of thermodynamics, which by mathematical representation is:

∆U= q + w. Where ∆U= change in internal enegy. Since the question is dealing with isothermal expansion, there is this rule that says for an isothermal expansion ∆U = 0.

Hence, 0 =q + [- 1717.3 Joules (J)].

q=1717.3 J = 1.7173 kJ.

Finally, the q(rev) which is= nRT ln (v2/V1).

q(rev) = 1 × 8.314 × 298 ln (2/1).

q(rev) = 1717.3 J = 1.7173 kJ.

PS: please note the negative signs in the workdone and the positive sign in the q(rev).

7 0

3 years ago

Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(

kkurt [141]

Answer: The $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)$ $\Delta H^o_{rxn}=?$