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77julia77 [94]
3 years ago
10

HELP!!!!!!!!!!!!!!!!!!! 100 POINTSSSSSSSSSSSSSS

Chemistry
2 answers:
e-lub [12.9K]3 years ago
7 0

Answer:

<em><u>The three-dimensional region of space that indicates where there is a high probability of finding an electron.</u></em>

jekas [21]3 years ago
7 0

Answer:

The orbital shown above is a 2s orbital. This is similar to a 1s orbital, except that the region where there is the greatest chance of finding the electron is further from the nucleus. This is an orbital at the second energy level.

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List earth and the gas giant planets from the hottest to the coldest planet
inessss [21]

Answer:

1. Venus

471°C

2. Mercury

(430°C) during the day,  (-180°C) at night

3. Earth

16°C

4. Mars

-28°C

5. Jupiter

-108°C

6. Saturn

-138°C

7. Uranus

-195°C

8. Neptune

-201°C

Explanation:

.

8 0
1 year ago
Zachary adds 26.64 g to 12.557 g. How many significant figures should his answer have?
asambeis [7]

This answer should have four signifigant features

Explanation:

I had this on a test and got it right  :D

5 0
3 years ago
Read 2 more answers
13. A gas occupies 4.31 liters at a pressure of 0.755 atm. Determine the volume if the
wariber [46]

Answer:

13. 2.60 L.

14. 2.40 L.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and T are constant, and have different values of P and V:

(P₁V₁) = (P₂V₂)

<em>13. A gas occupies 4.31 liters at a pressure of 0.755 atm. Determine the volume if the  pressure  is increased to 1.25 atm.</em>

P₁ = 0.755 atm, V₁ = 4.31 L.

P₂= 1.25 atm, V₂ = ??? L.

∴ V₂ = (P₁V₁)/(P₂) = (0.755 atm)(4.31 L)/(1.25 atm) = 2.60 L.

<em>14. 600.0 mL of a gas is at a pressure of 8.00 atm. What is the volume of the gas at 2.00  at m.</em>

P₁ = 8.0 atm, V₁ = 600.0 mL.

P₂= 2.0 atm, V₂ = ??? L.

∴ V₂ = (P₁V₁)/(P₂) = (8.0 atm)(600.0 mL)/(2.0 atm) = 2400/0 mL = 2.40 L.

3 0
2 years ago
11. The oxyfuel flame was used for fusion welding as early as the first half of the
Firlakuza [10]

The Oxyfuel gas or flame refers to a group of welding processes that use the flame produced by the combination of a fuel gas and oxygen as the source of heat.

<u>Explanation:</u>

  • Oxy-fuel welding is a process that utilizes fuel gases and oxygen to weld metals. Oxyfuel gas or flame refers to a group of welding processes that utilize the flame delivered by the blending of fuel gas and oxygen as the source of heat.
  • This flame is utilized for cutting and welding of two metallic pieces. This is done due to the heat produced by cutting and welding of two metallic pieces together by heating to the melting point.
  • An oxyhydrogen flame is utilized for cutting and welding of two metallic pieces due to the heat produced by the flame, i.e, 2800 ° C. At this temperature, the metal gets softened effectively and thus it can easily separate or welded together.
7 0
3 years ago
How many moles of HNO3 are present in 450 g of HNO3?
iVinArrow [24]
Since there are 63.01284 grams to one mole of HNO3, then 12.5 moles would be in 450 grams of it.
6 0
2 years ago
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