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2 years ago

What are some of the main features found on the Sun’s surface. Describe these features in detail.

1 answer:

7 0

Sun spots- dark circles on the sun’s surface that I believe are areas of cooler temperatures.
Solar flares- arcs of fire that leap from the sun’s surface. Nothing special, they’re a daily roumaine for the sun

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Calculate the pOH of this solution. pH = 1.90 pOH =

ladessa [460]

Answer:

Explanation:

Since the pH has been given, the pOH can be found by using the formula

pH + pOH = 14

pOH = 14 - pH

From the question we have

pOH = 14 - 1.9

We have the final answer as

Hope this helps you

8 0

2 years ago

Read 2 more answers

For the reaction Fe3O4(s) + 4H2(g) --> 3Fe(s) + 4H2O(g)

mojhsa [17]

Answer : The value of equilibrium constant for this reaction at 328.0 K is $1.70\times 10^{15}$

Explanation :

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy = 151.2 kJ = 151200 J

$\Delta S^o$ = standard entropy = 169.4 J/K

T = temperature of reaction = 328.0 K

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

$\Delta G^o=(151200J)-(328.0K\times 169.4J/K)$

$\Delta G^o=95636.8J=95.6kJ$

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln k$

where,

$\Delta G^o$ = standard Gibbs free energy = 95636.8 J

R = gas constant = 8.314 J/K.mol

T = temperature = 328.0 K

K = equilibrium constant = ?

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

$95636.8J=-(8.314J/K.mol)\times (328.0K)\times \ln k$

$k=1.70\times 10^{15}$

Therefore, the value of equilibrium constant for this reaction at 328.0 K is $1.70\times 10^{15}$

3 0

2 years ago

How many mole of H2 will be produced if 0.500 grams of magnesium is reacted with 10.00mL of 2.0 M HCl?

Ratling [72]

Yes, you're right the answer is 0,02 moles.

7 0

2 years ago

Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine. CCl4(g) + (1/2)O

madam [21]

The question is incomplete, here is the complete question:

Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine.

$CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c=4.4\times 10^9$ at 1,000 K

Calculate Kc for the reaction $2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)$

<u>Answer:</u> The value of $K_c'$ for the final reaction is $1.936\times 10^{19}$

<u>Explanation:</u>

The given chemical equations follows:

$CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c$

We need to calculate the equilibrium constant for the equation, which is:

$2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)$

As, the final reaction is the twice of the initial equation. So, the equilibrium constant for the final reaction will be the square of the initial equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c'=(K_c)^2$

We are given:

$K_c=4.4\times 10^9$

Putting values in above equation, we get:

$K_c'=(4.4\times 10^9)^2=1.936\times 10^{19}$

Hence, the value of $K_c'$ for the final reaction is $1.936\times 10^{19}$

3 0

2 years ago

4.2g of sodium bicarbonate is equivalent to how many moles of sodium bicarbonate

Mnenie [13.5K]

<span>The mass of one mole of sodium bicarbonate (aka NaHCO3) is equal to 1 * 22.99g/mol + 1 * 1.00g/mol + 1 * 12.01g/mol + 3 * 16.00g/mol = 83.91g/mol. From this, we can convert 4.2g of NaHCO3 to moles by dividing by 83.91g/mol, to get 0.050 moles of sodium bicarbonate.</span>

7 0

3 years ago

Read 2 more answers