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Dmitry_Shevchenko [17]
3 years ago
5

It get hotter towards the center

Chemistry
1 answer:
STatiana [176]3 years ago
8 0

Answer:

earth

Explanation:

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When bar-headed geese fly at very high altitudes (possibly over Mount Everest!), they breathe very thin air where the partial pr
Reil [10]

Answer:hemoglobin that has a high affinity for oxygen

Explanation:

Haemoglobin is the oxygen carrying pigment in blood. It performs this function because of the presence of iron at the center of the haemoglobin which coordinates reversibly with oxygen thereby aiding delivery of oxygen to cells. At high altitudes where air is thinner and the partial pressure of oxygen is lower than sea level, haemoglobin must develop a greater affinity for oxygen in order to carry the scarce oxygen to cells.

5 0
3 years ago
Noble gas notation write the electron configuration for the manganese atom
8_murik_8 [283]
<span> the electron configuration for the manganese atom [Ne]3s</span>²
6 0
3 years ago
What is the pOH of a<br> 5.6 x 10-5 M solution of cesium<br> hydroxide (CsOH)?
Rudiy27

Answer: 4.25

Explanation: CsOH=Cs+

pOH= -log10

= -log10(5.6*10^-5)

= -log10(5.6*10^-5)=-(-4.25)

=4.25

7 0
3 years ago
If a gas furnace releases 420 kcal of heat energy, what is the energy in kilojoules? (1 cal = 4.184 J)
Igoryamba
Answer:
The furnace releases 1757280 J 

Explanation:
We will do the conversion on two steps:
1- convert the kcal to cal
2- convert the cal to J

Step 1: converting kcal to cal
1 kcal is equivalent to 1000 cal. Therefore:
420 kcal is equivalent to 420*1000 = 420000 cal

Step 2: converting cal to J
We are given that:
<span>1 cal = 4.184 J
</span>Therefore:
420000 cal is equivalent to 420000 * 4.184 = 1757280 J 

Hope this helps :)
8 0
3 years ago
Consider the following system at equilibrium where H° = 111 kJ/mol, and Kc = 6.30, at 723 K.
Rashid [163]

Answer:

1) The value of Kc:

C. remains the same.

2) The value of Qc:

A. is greater than Kc.

3) The reaction must:

B. run in the reverse direction to restablish equilibrium.

4) The concentration of N2 will:

B. decrease.

Explanation:

Hello,

In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:

1) The value of Kc:

C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:

ln(K)=\frac{\Delta _RG}{RT}

2) The value of Qc:

A. is greater than Kc, since the reaction quotient is:

Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}

Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.

3) The reaction must:

B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.

4) The concentration of N2 will:

B. decrease, since less reactant is forming the products.

Best regards.

8 0
3 years ago
Read 2 more answers
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