Water boiling at 100° Celcius is it a <br><br> Chemical Property<br> Physical Property

As with any drug, aspirin (acetylsalicylic acid) must remain in the bloodstream long enough to be effective. Assume that the rem

Olenka [21]

Answer:

180 mg

Explanation:

For a first-order reaction, we can calculate the amount of aspirine (A) at a certain time (t) using the following expression.

$A=A_{0}.e^{-k.t}$

where,

k: rate constant

A₀: initial amount

If we know the half-life ( $t_{1/2}$ ) we can calculate the rate constant.

$k=\frac{ln2}{t_{1/2}} =\frac{ln2}{3h} =0.2h^{-1}$

When t = 4 h and A₀ = 400 mg, A is:

$A=400mg.e^{-0.2h^{-1}\times 4h} =180mg$

4 0

2 years ago

The electronegativity of nitrogen (n) is 3.0, while the electronegativity of hydrogen (h) is 2.1. knowing this, consider how the

Goryan [66]

The electronegativity of nitrogen (N) is 3.0, while the electronegativity of hydrogen (H) is 2.1. As it can be seen that nitrogen (N) is more electronegative than that of hydrogen (H),

So electron pairs are attracted towards nitrogen and thus it carries a partial negative charge and hydrogen carries a partial positive charge. The image of electron distribution is attached as follows.

Thus NH₃ is a polar molecule .

8 0

3 years ago

The electronic configuration of an element is given below.

Shalnov [3]

Answer:

It is reactive because it has to gain an electron to have a full outermost energy level.

Explanation:

The electron configuration of oxygen is 1s2,2s2 2p4.

Oxygen is in group six in the periodic table so it has six electrons in its valence shell. This means that it needs to gain two electrons to obey the octet rule and have a full outer shell of electrons (eight).

3 0

2 years ago

RATE LAW QUESTION !

vivado [14]

In general, we have this rate law express.:

$\mathrm{Rate} = k \cdot [A]^x [B]^y$
we need to find x and y

ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).

then we go to compare two experiments in which only one concentration is changed

compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4) by the smaller [B] (experiment 1) and call it Δ[B]

Δ[B]= 0.3 / 0.1 = 3

now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:

ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...

solve for y in the equation $\Delta \mathrm{Rate} = \Delta [B]^y$

$3.09 = (3)^y \implies y \approx 1$

To this point, $\mathrm{Rate} = k \cdot [A]^x [B]^1$

do the same to find x.
choose two experiments in which only the concentration of B is unchanged:

Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4

solve for x for $\Delta \mathrm{Rate} = \Delta [A]^x$

$4= (2)^x \implies x = 2$

the rate law is

Rate = k·[A]²[B]

6 0

2 years ago

Is gold a compound or an element

timurjin [86]

Gold is an element (Au) you can identify it on a periodic table

3 0

3 years ago

Water boiling at 100° Celcius is it a Chemical Property Physical Property