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nirvana33 [79]
3 years ago
5

Eroded soil carried off by water, wind, glaciers, or gravity is called __.

Chemistry
2 answers:
ankoles [38]3 years ago
5 0
A. Sediment

Hope this helps!
olga nikolaevna [1]3 years ago
5 0
A.Sediment is the answer
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How many grams of carbon dioxide will form if 5.5 g of C3H8 burns in 15 g of O2?
mr Goodwill [35]
C3H8+3O2--->3CO2+8H
Therefore for every 1:3 there are 3 Carbon dioxides that form. That means find the limiting reactant from the two reactants.
5.5g(1mole C3H8/44.03g of C3H8)=0.1249 moled of C3H8 and if for every one C3H8 we can form three CO2. We can assume 0.3747 miles of CO2 will be produced.
15g of O2(1 mole O2/32g of O2)=0.4685moles O2 and if for every three O2 we can produce three CO2 we may assume a 1:1 ratio.
This means C3H8 will be your limiting reactant. Therefore 0.3747 moles of CO2 will be produced.
0.3747 moles of CO2(48.01 g of CO2/1 mole of CO2)= 17.99 grams of CO2
5 0
3 years ago
How many lead atoms are in a cube of lead that has a volume of 2.18 cm3 if the density of lead is 11.3 g/cm3?
Anit [1.1K]
10000000.................
3 0
3 years ago
If the titrant has a molarity of 0.1000 m and there are 45.00 ml of analyte present, what is the molarity of the analyte?
marysya [2.9K]
Given: 

Concentration of titrant = 0.1000 M
Volume of titrant = 45 mL

The molarity of analyte depends on the amount of the analyte present in the titrated solution. If the amount of analyte is 20 mL, then its concentration is:

45ml * 0.10 M = C analyte * 20 ml

C analyte = 0.225 M
4 0
2 years ago
5. Write the names of the following ions.
Talja [164]
A.) Phosphate ion or Orthophosphate
d.) Hydroxide
D.) Ammonium
e.) Iron
C.) Nitrate
f.) Sulfur dioxide
5 0
2 years ago
Se construye una pila galvánica con una barra de cobre sumergida en una disolución 1M de cationes Fe+2 y una barra de plata sume
pashok25 [27]

Answer:

El potencial celular estándar, E_{cell} is +0.46 V

Explanation:

Las reacciones de media célula son;

Media reacción del ánodo Cu²⁺ + 2e⁻ ↔ Cu, E ° = 0.34 V

Media reacción catódica 2Ag + 2e⁻ ⁻ 2Ag, E ° = 0.80 V

Sin embargo tenemos para hierro Fe²⁺ + 2e⁻ ↔ Fe, E ° -0.44 V

y Fe³⁺ + e⁻ ↔ Fe²⁺, E ° = 0.77 V

 que es más alta que la del cobre presente, por lo tanto, el cobre se oxidará en el ánodo

Por lo tanto, en el ánodo, tendremos

Cu → Cu²⁺ + 2e⁻ (E ° = -0.34 V)

En el cátodo

2Ag + 2e⁻ → 2Ag (E ° = 0.80 V)

E_{cell} = E_c + E_a = -0.34 + 0.8 = +0.46 \, V

El potencial celular estándar, E_{cell} = +0.46 V

4 0
3 years ago
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