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timurjin [86]
3 years ago
13

1.

Chemistry
1 answer:
Alex_Xolod [135]3 years ago
3 0

i think that the answer is D

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I need help with these 4
storchak [24]

Rift Valley Examples

  • East African
  • Rhine Rift Valley (Germany)
  • Baikal Rift Valley (Russia)

Rift Valley Facts/Characteristics

  • Bordered by fault zones and separating land masses
  • A portion of land that drops, forming a valley.
  • Formed as tectonic plates move.

Fossils - Examples

  • Bones
  • Shells
  • Exoskeletons
  • Stone imprints of animals.

Non Examples of Fossils

  • Stones
  • rocks
  • Some bones (If they are fossils they will be a lot heavier than other bones.

Hope this helps.

5 0
3 years ago
How do seismologists, or scientists who study earthquakes, know that the largest known earthquake to hit the region occurred aro
denis23 [38]

Answer:

By looking at the amount of time between the P and S wave on a seismographs recorded on a seismograph, scientists can tell how far away the earthquake was from that location. However, they can't tell in what direction from the seismograph the earthquake was, only how far away it was.

Explanation:

5 0
3 years ago
Aluminum foil is often incorrectly termed tin foil. If the density of tin is 7.28g/cm³,what is the thickness of a piece of tin f
leonid [27]

Answer:

15.0 µm

Step-by-step explanation:

Density = mass/volume

        D = m/V     Multiply each side by V

      DV = m        Divide each side by D

        V = m/D

Data:

m = 1.091 g

D = 7.28 g/cm³

 l = 10.0 cm

w = 10.0 cm

Calculation:

<em>(a) Volume of foil </em>

V = 1.091 g  × (1 cm³/7.28 g)

  = 0.1499 cm³

(b) <em>Thickness of foil </em>

The foil is a rectangular solid.

V  = lwh                            Divide each side by lw

h = V/(lw)

   = 0.1499/(10 × 10)

   = 1.50 × 10⁻³ cm           Convert to millimetres

   = 0.015 mm                  Convert to micrometres

   = 15.0 µm

The foil is 15.0 µm thick.

4 0
3 years ago
_Al + _Fe(NO3)2 --&gt;_Fe + __Al(NO3)3
romanna [79]
4Al+6Fe(NO3)2= 6Fe+4Al(NO3)3
4 0
3 years ago
A compound made of two elements, iridium (Ir) and oxygen (O), was produced in a lab by heating iridium while exposed to air. The
natima [27]
1) Compund Ir (x) O(y)

2) Mass of iridium = mass of crucible and iridium - mass of crucible = 39.52 g - 38.26 g = 1.26 g

3) Mass of iridium oxide = mass of crucible and iridium oxide - mass of crucible = 39.73g - 38.26g = 1.47g

4) Mass of oxygen = mass of iridum oxide - mass of iridium = 1.47g - 1.26g = 0.21g

5) Convert grams to moles

moles of iridium = mass of iridium / molar mass of iridium = 1.26 g / 192.17 g/mol = 0.00656 moles

moles of oxygen = mass of oxygen / molar mass of oxygen = 0.21 g / 15.999 g/mol = 0.0131

6) Find the proportion of moles

Divide by the least of the number of moles, i.e. 0.00656

Ir: 0.00656 / 0.00656 = 1

O: 0.0131 / 0.00656 = 2

=> Empirical formula = Ir O2 (where 2 is the superscript for O)

Answer: Ir O2
5 0
4 years ago
Read 2 more answers
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