30 POINTS!!! Please Help With this!!!!

Why does copper give off blue light and lithium give off red light?

slava [35]

Because each element has an exactly defined line emission spectrum, scientists are able to identify them by the color of flame they produce.

5 0

3 years ago

The reaction 2 ClO2(g) + F2(g) → 2 FClO2(g) is first-order in both ClO2 and F2. When the initial concentrations of ClO2 and F2 a

kupik [55]

Answer:

3. 75.0%

Explanation:

2 ClO2(g) + F2(g) → 2 FClO2(g)

First order with respect to ClO2 and F2.

This means the rate equation is given as;

Rate = k [ClO2][F2]

When the initial concentrations of ClO2 and F2 are equal?

Let's assume an initial value of 1 for both reactants, so rate equation is given as;

Rate = k * 1 * 1 = k

The rate after 25% of the F2 has reacted is what percent of the initial rate?

The concentration left of F2 is 75% ( 100% - 25%) = 0.75

Concentration of ClO2 remains 1.

So rate equation is given as;

Rate = k * 1 * 0.75 = 0.75 k

Comparing 0.75k and k.

This means our answer is;

3. 75.0%

6 0

3 years ago

PLEASE HELP! WILL MARK BRAINIEST IF ANSWERED CORRECTLY.

ioda

Answer:

H2O is a polar molecule

the H is slightly positive and the O is slightly negative so the Na is attracted to the O side of the water molecule and the Cl is attracted to the H side of the water molecule

Explanation:

3 0

3 years ago

At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react

erastova [34]

Answer:

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

$N_{2 (g)}$ + $O_{2 (g)}$ ⇄ 2 $NO_{(g)}$

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no $N_{2 (g)}$ nor $O_{2 (g)}$ present. Immediately, $N_{2 (g)}$ and $O_{2 (g)}$ are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [ $N_{2 (g)}$ ]=0 ; [ $O_{2 (g)}$ ]= 0 ; [ $NO_{(g)}$ ]=0.60M

C: [ $N_{2 (g)}$ ]=+x ; [ $O_{2 (g)}$ ]= +x ; [ $NO_{(g)}$ ]=-2x

E: [ $N_{2 (g)}$ ]=0+x ; [ $O_{2 (g)}$ ]= 0+x ; [ $NO_{(g)}$ ]=0.60-2x

Now we can use the constant information:

$K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }$

$4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}$

$4.10* 10^{-4}$ = $\frac{(0.60-2x)^{2}}{x^{2} }$

$4.10* 10^{-4} * x^{2}$ = $(0.60-2x)^{2}}$

$\sqrt{4.10* 10^{-4} * x^{2}}$ = $\sqrt{(0.60-2x)^{2}}}$

$0.0202 x =0.60 - 2x$

$2x+0.0202x=0.60$

$x=\frac{0.60}{2.0202}$ $= 0.30$

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

3 0

2 years ago

The student ran out of time and did not do the second heating. Explain how this error would affect the calculation for the numbe

Ber [7]

The error caused will lower the number of moles when he did not do the second heating.This question is related to the empirical formula.

<h3>What is Empirical formula ?</h3>

A formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.

The empirical formula is the simplest ratio of atoms in a given compound.

we know that heating is used to eliminate most of the water molecules which are attached to the compound in the question, it's given that the second heating was not performed by the students.

This means that there are still some water molecules attached to the compound. This will imply that higher mass will be calculated for the compound because the water molecules are also considered.

So now, if we look at the calculation for the number of moles of water in this, we can say that this will be lower than the actual amount.

This is because there are some unaccounted moles of water which are still attached to the compound and these will be accounted in the compound mass rather than the waters. Therefore, the number of moles will be lower.

Hence, The error caused will lower the number of moles when he did not do the second heating.

Learn more about Empirical formula here ;

https://brainly.in/question/32699671

#SPJ1

6 0

2 years ago