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creativ13 [48]
3 years ago
8

Last one says hydrosphere you can’t see it though help please really fats please

Chemistry
1 answer:
DaniilM [7]3 years ago
3 0

Answer:

c

Explanation:

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A scientist measures the standard enthalpy change for the following reaction to be -327.2 kJ : P4O10(s) 6 H2O(l)4H3PO4(aq) Based
OleMash [197]

Answer:

ΔH°f P4O10(s) = - 3115.795 KJ/mol

Explanation:

  • P4O10(s) + 6H2O(l) ↔ 4H3PO4(aq)
  • ΔH°rxn = ∑νiΔH°fi

∴ ΔH°rxn = - 327.2 KJ

∴ ΔH°f H2O(l) = - 285.84 KJ/mol

∴ ΔH°F H3PO4(aq) = - 1289.5088 KJ/mol

⇒ ΔH°rxn = (4)(- 1289.5088) - (6)(- 285.84) - ΔH°f P4O10(s) = - 327.2 KJ

⇒ ΔH°f P4O10(s) = - 5158.035 + 1715.04 + 327.2

⇒ ΔH°f P4O10(s) = - 3115.795 KJ/mol

5 0
4 years ago
In the mass spectrum of the molecule phenol, C6H5OH, the approximate intensity of the peak at m/z 95, relative to the molecular
klasskru [66]
The correct answer to thi
6 0
4 years ago
Explain how the tasks an organism completes will influence their cellular respiration levels.
Debora [2.8K]

Answer:

All living things use cellular respiration to turn organic molecules into energy. ... This process makes energy from food molecules available for the organism to carry out life processes. Cellular respiration usually occurs in the presence of oxygen. This is called aerobic respiration.

6 0
3 years ago
At 350 k, kc = 0.142 for the reaction 2 brcl(g) ⇀↽ br2(g) + cl2(g) an equilibrium mixture at this temperature contains equal con
djyliett [7]
0.114 mol/l  
The equilibrium equation will be: 
Kc = ([Br2][Cl2])/[BrCl]^2  
The square factor for BrCl is due to the 2 coefficient on that side of the equation.  
Now solve for BrCl, substitute the known values and calculate. 
Kc = ([Br2][Cl2])/[BrCl]^2 
[BrCl]^2 * Kc = ([Br2][Cl2]) 
[BrCl]^2 = ([Br2][Cl2])/Kc 
[BrCl] = sqrt(([Br2][Cl2])/Kc)  
[BrCl] = sqrt(0.043 mol/l * 0.043 mol/l / 0.142) 
[BrCl] = sqrt(0.001849 mol^2/l^2 / 0.142) 
[BrCl] = sqrt(0.013021127 mol^2/l^2) 
[BrCl] = 0.114110152 mol/l  
Rounding to 3 significant figures gives 0.114 mol/l
4 0
3 years ago
1.Calculate the molarity of a 6.15% (w/v%) sodium hypochlorite (NaOCl) solution (i.e. in a 100 ml of solution there is 6.15 gram
shtirl [24]

Answer:

C  U  M

Explanation:

3 0
3 years ago
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