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GREYUIT [131]
3 years ago
13

_______A representative element in the fourth period that is an alkaline earth

Chemistry
1 answer:
MariettaO [177]3 years ago
4 0

Answer:calcium

Explanation: calcium is an alkaline earth metal,group 2 period 4

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Particles of clay and silt eroded and deposited by the wind are called
Ivanshal [37]
The correct answer to your question would be B) Loess, or <span>Aeolian deposits , reasoning to your question is because, loess is a german word but in english means loss or loose. So given that clay and small particles that are not combined together are loose particles. moves freely. Hope this helps you out.

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4 0
3 years ago
Read 2 more answers
Can someone answer the bottom question (22-25)
Sindrei [870]
22) Oxygen with a -2 charge
23) Neon with a +2 charge
24) Fluorine with a -1 charge
25) Lithium with a +1 charge
8 0
2 years ago
The Goodyear blimp contains 5.7 x 10^6 L of helium at 25 degrees Celsius and 1 atm. What is the mass in grams of the helium insi
anygoal [31]

Answer:

1.72x10⁻⁵ g

Explanation:

To solve this problem we use the PV=nRT equation, where:

  • P = 1 atm
  • V = 5.7x10⁶ L
  • n = ?
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • T = 25 °C ⇒ (25+273.16) = 298.16 K

And we <u>solve for n</u>:

  • 1 atm * 5.7x10⁶ L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K
  • n = 4.29x10⁻⁶ mol

Finally we <u>convert moles of helium to grams</u>, using its <em>molar mass</em>:

  • 4.29x10⁻⁶ mol * 4 g/mol = 1.72x10⁻⁵ g

8 0
2 years ago
An atomic spectra is like a _________ for an element, in that it is unique for that element and can be used in identification.
adell [148]
Atomic number, because the atomic number is unque to each individua element.
3 0
3 years ago
how many atoms are contained in 2.70g of aluminum provided that 32g of sulphur equals 6.02 × 10^(23)atoms​
umka2103 [35]

Answer:

1.63 \times  {10}^{24}

one atom of an element = 6.02 \times {10}^{23} atom

The mass of one atom of sulphur = 32g

The mass of one atom of aluminium = 27g

so one atom of aluminium = 6.02 \times {10}^{23}

27g of AL = 6.02 \times {10}^{23} atom

2.70g of AL = X atoms

Then you cross multiply ........

and get the answer

.

4 0
2 years ago
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