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Zigmanuir [339]
3 years ago
14

Please help me with number 43

Chemistry
1 answer:
Lana71 [14]3 years ago
8 0

Answer:

The reason why atomic mass is usually not a whole number is because it is a weighted average of the mass numbers of isotopes

Explanation:

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Which coefficient before potassium (K) balances this chemical equation?
miskamm [114]

Answer:

The coefficient before potassium (K) balances this chemical equation is 2.

Explanation:

_K +Cl₂ → 2KCl

K =1 ; Cl =2

K=1 × 2 = 2

Cl = 1 × 2 = 2

2 K +Cl₂ = 2 KCl

5 0
3 years ago
Read 2 more answers
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
3 years ago
Which compound is responsible for the smell that is associated with the decomposition of dead animals?
CaHeK987 [17]
the results from the breakdown of proteins and amino acids causing a foul smell
3 0
3 years ago
Calculate the mass in grams of each sample.<br> 4.88x10^20 H2O2 molecules
AleksAgata [21]

4.88x10^20 H2O2 molecules

5 0
3 years ago
A bubble of helium gas has a volume of 0.65 L near the bottom of a large
Flura [38]

Answer:

0.73L

Explanation:

The following data were obtained from the question :

V1 = 0.65 L

P1 = 3.4 atm

T1 = 19°C = 19 + 273 = 292K

V2 =?

P2 = 3.2 atm

T2 = 36°C = 36 + 273 = 309K

The bubble's volume near the top can be obtain as follows:

P1V1 /T1 = P2V2 /T2

3.4 x 0.65/292 = 3.2 x V2 /309

Cross multiply to express in linear form as shown below:

292 x 3.2 x V2 = 3.4 x 0.65 x 309

Divide both side by 292 x 3.2

V2 = (3.4 x 0.65 x 309) /(292 x 3.2)

V2 = 0.73L

Therefore, the bubble's volume near the top is 0.73L

8 0
3 years ago
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