What is a unique quality of the halogens/halides?

rusak2 [61]

Answer; molecule

Explanation:

5 0

3 years ago

Chlorine has two stable isotopes, Cl-35 and Cl-37. If their exact masses are 34.9689 amu and 36.9695 amu, respectively, what is

natima [27]

Answer:

X(Cl-35) = 75.95% => Answer 'A'

Explanation:

34.9689·X(Cl-35) + 36.9695·X(Cl-37) = 35.45; X = fractional abundance

X(Cl-35) + X(Cl-37) = 1 ⇒ X(Cl-37) = 1 - X(Cl-25)

34.9689·X(Cl-35) + 36.9695(1 - X(Cl-35)) = 35.45

34.9689·X(Cl-35) + 36.9695 - 36.9695·X(Cl-35) = 35.45

Rearrange ...

36.9695·X(Cl-35) - 34.9689·X(Cl-35) = 36.9689 - 35.45

2.0006·X(Cl-35) = 1.5195

X(Cl-35) = 1.5195/2.0006 = 0.7595 fractional abundance

⇒ % abundance = 75.95%

3 0

3 years ago

Read 2 more answers

Two 10g blocks, one of copper and one of iron, were heated from 300 K to 400K (a temperature difference of 100 K).

vova2212 [387]

Energy absorbed by Iron block E (iron) = 460.5 J

Energy absorbed by Copper block E (Copper) = 376.8 J

<u>Explanation:</u>

To find the heat absorbed, we can use the formula as,

q = m c ΔT

Here, Mass = m = 10 g = 0.01 kg

ΔT = change in temperature = 400 - 300 = 100 K = 100 - 273 = -173 °C

c = specific heat capacity

c for iron = 460.5 J/kg K

c for copper = 376.8 J/kg K

Plugin the values in the above equation, we will get,

q (iron) = 0.01 kg × 460.5 J/kg K × 100 K

= 460.5 J

q (copper) = 0.01 kg × 376.8 J/kg K × 100 K

= 376.8 J

3 0

3 years ago

How much 2.0 M NH4NO3 is needed to make 0.585 L of 1.2 M NH4NO3 solution?

Nutka1998 [239]

Answer:

b . 0.351 L.

Explanation:

Hello!

In this case, since diluted solutions are prepared by adding an extra amount of diluent to a stock-concentrated solution, we infer that the number of moles of solute remains the same, therefore we can write:

$C_1V_1=C_2V_2$

Thus, solving for the volume of the stock solution, V1, we obtain:

$V_1=\frac{C_2V_2}{C_1}$

Now, by plugging in the given data we obtain:

$V_1=\frac{1.2M*0.585L}{2.0M}\\\\V_1=0.351L$

Therefore, the answer is b . 0.351 L.

Best regards!

5 0

2 years ago

Ill give the brainliest answer to whoever helps me with this equation

vampirchik [111]

Answer: The percent yield for the $NaBr$ is, 86.7 %

Explanation : Given,

Moles of $FeBr_3$ = 2.36 mol

Moles of $NaBr$ = 6.14 mol

First we have to calculate the moles of $NaBr$

The balanced chemical equation is:

$2FeBr_3+3Na_2S\rightarrow Fe_2S_3+6NaBr$

From the reaction, we conclude that

As, 2 moles of $FeBr_3$ react to give 6 moles of $NaBr$

So, 2.36 moles of $FeBr_3$ react to give $\frac{6}{2}\times 2.36=7.08$ mole of $NaBr$

Now we have to calculate the percent yield for the $NaBr$ .

$\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield = 6.14 moles

Theoretical yield = 7.08 moles

Now put all the given values in this formula, we get:

$\text{Percent yield}=\frac{6.14mol}{7.08mol}\times 100=86.7\%$

Therefore, the percent yield for the $NaBr$ is, 86.7 %

6 0

2 years ago