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Strike441 [17]
3 years ago
11

Help plzzzzzzz ASAP!

Chemistry
2 answers:
Agata [3.3K]3 years ago
8 0
I am going to say C. it has to do with the angles
kaheart [24]3 years ago
3 0
The index of refraction is the speed of light in a material so the answer is c

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Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir
guajiro [1.7K]

Answer:

y_{O2} =4.3%

Explanation:

The ethanol combustion reaction is:

C_{2}H_{5} OH+3O_{2}→2CO_{2}+3H_{2}O

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}

Dividing the previous equation by x:

1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles

Calculate the number of moles of CO2 and water considering the same:

0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)

0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)

The total number of moles at the reactor output would be:

N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)

So, the oxygen mole fraction would be:

y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3%

6 0
3 years ago
Express as ordinary numbers. 3 x 100 =
Stolb23 [73]
The answer is 300
3 * 100
3 * 10 * 10
30 * 10
300

Hope this helps. If it did, consider making it Brainliest!
7 0
2 years ago
Read 2 more answers
Describe the region in model 1 where the highest concentration of hydrogen ion (h + ) is located
neonofarm [45]
<span>THE HIGHEST CONCENTRATION OF HYDROGEN IONS IS LOCATED IN THE INTER-MEMBRANE SPACE. HYDROGEN IONS REACH THE INTER-MEMBRANE SPACE THROUGH PROTEIN CHANNELS EMBEDDED IN THE MITOCHONDRIAL MEMBRANE. THE MAIN FUNCTION OF INTER MEMBRANEIS OXIDATIVE PHOSPHORLATON. ENERGY IS REQUIRED TO MOVE THE HYDROGEN IONS ACROSS THE MEMBRANE BECAUSE THE HYDROGEN IONS ARE MOVING AGAINST THE CONCENTRATION GRADIENT. H+ GOES AGAINST THE CONCENTRATION GRADIENT THE USE OF THE GRADIENT TO DRIVE ATP SYNTHASE. HYDOGEN IONS DRIVE ATP SYNTHASE IN PHTOSYNTHESIS. THIS HAPPENS WHEN HYDROGEN IONS GET PUSHED ACROSS THE MEMBRANE CREATING A HIGH CONCENTRATION INSIDE THE THYLAKOID.</span>
7 0
3 years ago
A solution of silver chlorate reacts with a solution of lithium iodide.
Anettt [7]

Answer: Ag^+(aq)+I^-(aq)\rightarrow AgI(s)

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.  

Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.

The given chemical equation is:

AgClO_3(aq)+LiI(aq)\rightarrow AgI(s)+LiClO_3(aq)

The complete ionic equation is:

Ag^+(aq)+ClO_3^-(aq)+Li^+(aq)+I^-(aq)\rightarrow AgI(s)+Li^+(aq)+ClO_3^-(aq)

The ions which are present on both the sides of the equation are lithium and chlorate ions. and hence are not involved in net ionic equation.

Thus the net ionic equation is:

Ag^+(aq)+I^-(aq)\rightarrow AgI(s)

7 0
2 years ago
A metal ion Mⁿ⁺ has a single electron. The highest energy line in its emission spectrum occurs at a frequency of 2.961 x 10¹⁶ Hz
Whitepunk [10]

Answer:

z≅3

Atomic number is 3, So ion is Lithium ion (Li^+)

Explanation:

First of all

v=f*λ

In our case v=c

c=f*λ

λ=c/f

where:

c is the speed of light

f is the frequency

\lambda=\frac{3*10^8}{2.961*10^{16}}\\ \lambda=1.01317*10^{-8} m

Using Rydberg's Formula:

\frac{1}{\lambda}=R*z^2*(\frac{1}{n_1^2}-\frac{1}{n_2^2})

Where:

R is Rydberg constant=1.097*10^7

z is atomic Number

For highest Energy:

n_1=1

n_2=∞

\frac{1}{1.01317*10^{-8}}=1.097*10^{7}*z^2*(\frac{1}{1^2}-\frac{1}{\inf})\\z^2=8.99\\z=2.99

z≅3

Atomic number is 3, So ion is Lithium ion (Li^+)

3 0
3 years ago
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