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Zarrin [17]
3 years ago
15

Which glacial landforms require more than one glacier to form

Chemistry
2 answers:
ruslelena [56]3 years ago
7 0

Answer:

C. horns, arêtes, drumlins, and hanging valleys

grandymaker [24]3 years ago
5 0
<span>aretes,hanging valleys,horns,and drumlins here you go!</span>
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I need to know the weighted average please!
maria [59]

Answer:

63.5456 amu

Explanation:

0.6917*62.9296+0.3083*64.9278= 63.5456\ amu

8 0
3 years ago
What volume would a sample of gas occupy in LITERS at 0.985 atmospheres and a volume of 3.65 liters if the pressure were raised
musickatia [10]

Answer:

3.18 L

Explanation:

Step 1: Given data

  • Initial pressure (P₁): 0.985 atm
  • Initial volume (V₁): 3.65 L
  • Final pressure (P₂): 861.0 mmHg
  • Final volume (V₂): ?

Step 2: Convert P₁ to mmHg

We will use the conversion factor 1 atm = 760 mmHg.

0.985 atm × 760 mmHg/1 atm = 749 mmHg

Step 3: Calculate the final volume of the gas

Assuming ideal behavior and constant temperature, we can calculate the final volume using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 749 mmHg × 3.65 L/861.0 mmHg = 3.18 L

7 0
3 years ago
What is the term for a bond composed of three electron pairs shared between two atoms?
liraira [26]
A covalent bond? Not sure how much detail you want, sorry
7 0
3 years ago
0.415 g of an unknown triprotic acid are used to make a 100.00 mL solution. Then 25.00 mL of this solution is transferred to an
Arturiano [62]

Answer:

Explanation:

Initial burette reading = 1.81 mL

final burette reading = 39.7 mL

volume of NaOH used = 39.7 - 1.81 = 37.89 mL .  

37.89 mL of .1029 M NaOH is used to neutralise triprotic acid

No of moles contained by 37.89 mL of .1029 M NaOH

= .03789 x .1029 moles

= 3.89 x 10⁻³ moles

Since acid is triprotic ,  its equivalent weight = molecular weight / 3

No of moles of triprotic acid = 3.89 x 10⁻³ / 3

= 1.30   x 10⁻³ moles .

8 0
3 years ago
The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How
maks197457 [2]
2iekfjdjabnfwnfuwj ag hbn few nbfnbfb
4 0
3 years ago
Read 2 more answers
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