What percentage of the sun is composed of hydrogen?

ruslelena [56]

Answer:

O2, oxygen.

Explanation:

Hello.

In this case, for the undergoing chemical reaction, we need to compute the moles of CO2 yielded by 85 g of CH4 (molar mass = 16 g/mol) and by 320 g of O2 (molar mass 32 g/mol) via the following mole-mass relationships:

$n_{CO_2}^{by\ CH_4}=85gCH_4*\frac{1molCH_4}{16gCH_4} *\frac{1molCO_2}{1molCH_4} =5.3molCO_2\\\\n_{CO_2}^{by\ O_2}=320gO_2*\frac{1molO_2}{32gO_2} *\frac{1molCO_2}{2molO_2} =5molCO_2$

Considering the 1:2:1 among CH4, O2 and CO2. Therefore, since 320 g of O2 yield the smallest amount of CO2 we infer that the limiting reactant is O2.

Best regards.

5 0

3 years ago

Calculate the mass, in grams, of Al(OH)3 required for this reaction:

AVprozaik [17]

Answer:

Mass = 28.08

Explanation:

Given data:

Mass of Al₂O₃ = 21.8 g

Mass of water = 9.7 g

Mass of Al(OH)₃ = ?

Solution:

Chemical equation:

2Al(OH)₃ → Al₂O₃ + 3H₂O

Number of moles of water:

Number of moles = mass/molar mass

Number of moles = 9.7 g/ 18 g/mol

Number of moles = 0.54 mol

Number of moles of Al₂O₃:

Number of moles = mass/molar mass

Number of moles = 21.8 g/ 101.96 g/mol

Number of moles = 0.21 mol

Now we will compare the moles of Al(OH)₃ with Al₂O₃ and H₂O.

Al₂O₃ : Al(OH)₃

1 : 2

0.21 : 2×0.21 = 0.42 mol

H₂O : Al(OH)₃

3 ; 2

0.54 : 2/3×0.54 = 0.36 mol

Mass of Al(OH)₃:

Mass = number of moles × molar mass

Mass = 0.36 mol × 78 g/mol

Mass = 28.08 g

3 0

3 years ago

Nitrogen and oxygen do not react appreciably at room temperature, as illustrated by our atmosphere. But at high temperatures, th

Gre4nikov [31]

Answer : The equilibrium concentration of NO is, 0.0092 M.

Solution :

First we have to calculate the concentration of NO.

$\text{Concentration of NO}=\frac{\text{Moles of }NO}{\text{Volume of solution}}=\frac{0.3152mol}{2.0L}=0.1576M$

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

Initially conc. 0 0 0.1576

At eqm. (x) (x) (0.1576-2x)

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$0.0153=\frac{(0.1576-2x)^2}{(x)\times (x)}$

By solving the term, we get:

$x=0.0742,0.0839$

Neglecting the 0.0839 value of x because it can not be more than initial value.

Thus, the value of 'x' will be, 0.0742 M

Now we have to calculate the equilibrium concentration of NO.

Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M

Therefore, the equilibrium concentration of NO is, 0.0092 M.

5 0

4 years ago

Please help will mark brainliest!! What is the main point of paragraph 3?

zmey [24]

Answer:

the highest frequency

Explanation:

8 0

3 years ago

Can someone please do this for me

Setler [38]

2K(s) + F₂(g) --> 2KF(s)

The physical stare of potassium is solid(s) and that of potassium fluoride is solid(s) so, s will be put on both the blanks.

Now, In order to balance florine, we will put 2 for potassium floride, afterwards florone will be 2 on both side of equation but now potasium will become unbalnced so, we will put 2 for potassium. Thus, both sides of the equation will be balanced.

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P.S- Sorry, If I fail in properly explaining it to you.

8 0

3 years ago