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gizmo_the_mogwai [7]
3 years ago
5

What if you mix green with blue?

Chemistry
1 answer:
slamgirl [31]3 years ago
6 0

Answer:

Ooo points, You get teal, turquoise, and cyan :)

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Which of the following combination of elements would result in covalent compound? * W X Y Z Vand X Wand Z Y and Z Wand y​
TEA [102]

Answer:

C. Y & Z

Explanation:

V, W are imaginary metals here because their valence electrons are typically less than 4. X, Y, Z are non-metals and have higher valence electrons. Here, if V or W bind with X, Y, or Z we make ionic bond (because metal + non metal = ionic). But, if X binds with Y or Z or any combinations of any two of the three non-metals results in covalent bond (non metal + non metal = covalent).

Thus, Y and Z make covalent.

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3 years ago
Orygon
grigory [225]

if 105 grams burns completely

therefore

105 ×22.4/48=49

8 0
3 years ago
What is the function
Verizon [17]

Answer:

c. carry materials to all parts of the body

Explanation:

The lungs and tissues receive oxygen and nourishment via blood.

8 0
3 years ago
en un recipiente se tiene 800 g de una solución al 35% en masa de ácido sulfuroso, de la cual se evapora 80ml de agua. ¿cuál es
Alinara [238K]

The mass percent of sulfurous acid in the new solution : 38.9%

<h3>Further explanation</h3>

<em>In a container you have 800 g of a 35% by mass solution of sulfurous acid, from which 80 ml of water evaporates. What is the mass percent of sulfurous acid in the new solution? data: density of water is 1g / ml.</em>

<em />

solution 1

composition :

  • 35% acid :

\tt 0.35\times 800~g=280~g

  • water :

\tt 800-280=520~g

solution 2(new solution)

composition :

  • water

\tt 520-(80~ml\times 1~g/ml)=440~g

  • Total mass of new solution after water evaporated

\tt 280(acid)+440(water)=720~g

  • %mass of acid in a new solution

\tt \dfrac{280}{720}\times 100\%=38.9\%

5 0
3 years ago
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
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