The bathroom mirror has water droplets on it after a hot shower. choose the correct phase change. HELPP

The colour of light depends on it's wave length.Red light have a wave length in the order of 7.8*10 raise to power of -7metre, e

Arisa [49]

That will be 0.78*10 raise to the power of -1 nanometers

8 0

3 years ago

Ibuprofen is a common pain reliever and anti-inflammatory. Its formula is C13H18O2. What percent by mass of ibuprofen is the car

Anarel [89]

Answer:

Explanation:

%Carbon = mass of carbon / mas of C13H18O2) X100%

= 13X12/ (13X12+18+2X16)X100%

= 156/206) X100%= 75.7%

8 0

3 years ago

The reaction of sodium peroxide and water produces sodium hydroxide and oxygen gas. The following balanced chemical equation rep

a_sh-v [17]

Answer:

3.925 mol.

Explanation:

From the balanced equation:

2 Na₂O₂(s) + 2 H₂O(l) → 4 NaOH(s) + O₂(g) ,

It is clear that 2 moles of Na₂O₂ react with 2 moles of H₂O to produce 4 moles of NaOH and 1 mole of O₂ .

Using cross multiplication:

4 moles of NaOH produced with → 1 mole of O₂ .

15.7 moles of NaOH produced with → ??? mole of O₂ .

∴ The no. of moles of O₂ made = (1 mole)(15.7 mole)/(4 mole) = 3.925 mol.

8 0

3 years ago

Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has

Ymorist [56]

Answer:

$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

Explanation:

Given that:

The Half-life of $^{235}U$ = $7.13 \times 10^8 \ years$ is less than that of $^{238} U = 4.51 \times 10^9 \ years$

Although we are not given any value about the present weight of $^{235}U$ .

So, consider the present weight in the percentage of $^{235}U$ to be y%

Then, the time elapsed to get the present weight of $^{235}U$ = $t_1$

Therefore;

$N_1 = N_o e^{-\lambda \ t_1}$

here;

$N_1$ = Number of radioactive atoms relating to the weight of y of $^{235}U$

Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

$In( \dfrac{N_o}{N_1}) = \lambda t_1$ --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of $^{235}U$ to be = $t_2$

Then:

$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

here;

$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

replacing the half-life of $^{235}U$ = $7.13 \times 10^8 \ years$

$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

∴

$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is $\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

8 0

3 years ago

WHAT IS THE CHEMICAL NAME FOR ALKYNE HAVING

madreJ [45]

Answer:

Ee your house 6r2f5r56rrrr6gjyf

Explanation:

4 0

2 years ago

The bathroom mirror has water droplets on it after a hot shower. choose the correct phase change￼. HELPP

The bathroom mirror has water droplets on it after a hot shower. choose the correct phase change. HELPP