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3 years ago

True or False

Chemistry

2 answers:

daser333 [38]3 years ago

8 0

Answer:

if its both opposite symbols, they will attract, if they are the same symbols, they will repel

Explanation:

Hope it helps ( :

(P.S. there's no pic)

Softa [21]3 years ago

7 0

Answer:

True

Explanation:

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How does the density of a hershey bar compare to the density of the a half bar?

anastassius [24]

The density would be the same for the whole bar as well as one half of the bar. Density is a identity I believe, by this I mean that it stays the same no matter how little or how much of the same substance you have. Since density = mass / volume, half the bar has half of the weight as well as half of the volume of the whole bar, making the density the same.

For example, a block weighs 10 grams and has a volume of 5 ml. the density would be d = 10/5 or, d = 2g/ml

Half of the block weighs 5 grams and has a volume of 2.5 ml. The density is d = 5/2.5, or, d = 2 g/ml.

See, although there are different amounts of the same substance, their density is the same.

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3 years ago

What is the mass in grams of 4.25 x 103moles of N2?

Dovator [93]

Mass =?
moles of N2 = 4.25 x 103 mol
molar mass of N2 = (14)x2 = 28

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3 years ago

Zinc reacts with excess hydrochloric acid to produce zinc chloride and hydrogen gas. What happens to the reaction rate if you in

agasfer [191]

The correct response I believe is D. The reaction rate increases because the probability of collisions increases as there are more Zn atoms to react.

4 0

3 years ago

Read 2 more answers

In atmospheric chemistry, the following chemical reaction converts SO2, the predominant oxide of sulfur that comes from combusti

Misha Larkins [42]

Answer:

Explanation:

From the given information;

The chemical reaction can be well presented as follows:

$\mathtt{SO_{2(g)} + \dfrac{1}{2}O_{2(g)} }$ ⇄ $\mathtt{3SO_{2(l)}}$

Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:

i.e

$K = \dfrac{a_{so_3}}{a_{so_2} a_{o_2}^{\frac{1}{2}}}$

However, since we are dealing with liquids solutions;

$K = \dfrac{1}{\dfrac{Pso_2}{P^0}\Big ( \dfrac{Po_2}{P^0} \Big)^{1/2}}$ since the activity of $a_{so_3}$ is equivalent to 1

Hence, under standard conditions(i.e at a pressure of 1 bar)

$K = \dfrac{1}{Pso_2Po_2^{1/2}}$

(b)

From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:

$\Delta _{rxn} G^o = \sum \Delta_f \ G^o (products) - \sum \Delta_fG^o (reactants) \\ \\ = (1) (-368 \ kJ/mol) - (\dfrac{1}{2}) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\ \simeq -68 \ kJ/mol$

Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol

(c)

Le's recall that:

At equilibrium, the instantaneous free energy is usually zero &

Q(reaction quotient) is equivalent to K(equilibrium constant)

So;

$\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}$

$\mathtt{0- \Delta _{rxn} G^o = RTIn K } \\ \\ \mathtt{ \Delta _{rxn} G^o = -RTIn K } \\ \\ K = e^{\dfrac{\Delta_{rxn} G^o}{RT}} \\ \\ K = e^{^{\dfrac{67900 \ J/mol}{8.314 \ J/mol \times 298 \ K}} }$

$K =7.98390356\times 10^{11} \\ \\ \mathbf{K = 7.98 \times 10^{11}}$

(d)

The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).

This is because;

If Q < K, then the reaction will proceed in the right direction towards the products.

However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.

So;

$Q= \dfrac{1}{Pso_2Po_2^{1/2}}$

Since we are dealing with liquids;

$Q= \dfrac{1}{1 \times 1^{1/2}}$

Q = 1

Since Q < K; Then, the reaction proceeds in the right direction.

Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.

8 0

3 years ago

Using words, explain all steps of dimensional analysis when converting 6.7 hectometers to meters. You must explain the set up of

sleet_krkn [62]

Answer:

See below

Explanation:

In order to convert 6.7 hectometers to meters using dimensional analysis, you must use the ratio of hectometers to meters.

There are 100 meters in one hectometer, then you start setting the equality: 100 m = 1 hm.

Dividing both sides by 1 hm you get the fraction form of the ratio, with meters (m) in the numerator and hectometers (hm) in the denominator.

$1m/100hm$

Since hm is in the numerator, when you multiply 6.7 hm by the unit ratio the hm will be canceled and the result will have only m:

$6.7hm\times 1m/100hm=0.067m$

6 0

3 years ago