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laila [671]
3 years ago
14

During the hot summer days, there is a lot more water in the air. This is due to which change of state?

Chemistry
2 answers:
Alex73 [517]3 years ago
4 0
The answer is B evaporation.
Bond [772]3 years ago
4 0

B) Evaporation

Hope you got it right :)

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Atoms with_____or less electrons in the outermost level tend to lose electrons
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Atoms with<u>2</u>or less electrons in the outermost level tend to lose electrons

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2 years ago
You carefully weigh out 16.00 g of CaCO3 powder and add it to 64.80 g of HCl solution. You notice bubbles as a reaction takes pl
bulgar [2K]

Answer: Mass of CO_2  produced in this reaction was 6.56 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

CaCO_3(s)+2HCl(aq)\rightarrow H_2O(l)+CO_2(g)+CaCl_2(aq)

Mass or reactants =  Mass of CaCO_3+ mass of HCl = 16.00 + 64.80 = 80.80 g

Mass of products  = mass of aqueous solution + mass of CO_2 + = 74.24 + x g

Mass or reactants = Mass of products

80.80 g = 74.24 + x g

x = 6.56 g

Thus mass of CO_2  produced in this reaction was 6.56 grams

7 0
3 years ago
What type of animal might become preserved in amber
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Insect's or Bugs preserved in Amber
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The standard enthalpy of formation for glucose [c6h12o6(s)] is −1273.3 kj/mol. what is the correct formation equation correspond
balu736 [363]
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is 
     C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is 
     6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)

Using the equation for the standard enthalpy change of formation 
     ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
     ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}

C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
     ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
                           = -1273.3 - (0 + 0 + 0)
                           = -1273.3
8 0
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