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Anvisha [2.4K]
2 years ago
15

If you are traveling at 60mi/hr, how many meters per second is that equivalent to?

Chemistry
1 answer:
lesya [120]2 years ago
5 0

Answer:

26.8224 Meters per Second.      

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Based on the information that is given, which atom is the table ?
andriy [413]

Answer:

The one with the greatest mass would be the one that has the most things in the nucleus, protons and nutrons

Explanation:

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3 years ago
Place where food is stored and partially digested
Ostrovityanka [42]
Answer: Chyme

Explain: Chyme passes from the stomach to the small intestine. Further protein digestion takes place in the small intestine.
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3 years ago
Which statement best describes how Carl Woese changed the system of classification?
Nadya [2.5K]

Answer: C.)

Explanation:

i got it right on a unit test!

but it might be something else if there arranged different!

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8 0
3 years ago
Read 2 more answers
Brainliest for an answer!
Degger [83]

The volume of H₂ : = 15.2208 L

<h3>Further explanation</h3>

Given

Reaction

2 As (s) + 6 NaOH (aq) → 2 Na₃AsO₃ (s) + 3 H₂ (g)

34.0g of As

Required

The volume of H₂ at STP

Solution

mol As (Ar = 75 g/mol) :

= mass : Ar

= 34 g : 75 g/mol

= 0.453 mol

From the equation, mol ratio As : H₂ = 2 : 3, so mol H₂ :

=3/2 x mol As

=3/2 x 0.453

= 0.6795

At STP, 1 mol = 22.4 L, so :

= 0.6795 x 22.4 L

= 15.2208 L

5 0
3 years ago
En un matraz, disponemos de 100 g de gas oxígeno que se encuentran a 1 at de presión y 273 K de temperatura. Calcular : a) el nú
Misha Larkins [42]

Answer:

Explanation:

Dado que:

masa de oxígeno gaseoso = 100 g

presión = 1 atm

temperatura = 273 K

(a)

número de moles de oxígeno contenidos en el matraz = masa de oxígeno / masa molar de oxígeno

= 100 g / 16 gmol⁻¹

= 6.25 moles

(b) El número de moléculas de oxígeno es el siguiente:

Dado que 1 mol de oxígeno gaseoso contiene 6.023 * 10²³ moléculas de oxígeno.

Entonces, 6.25 moles contendrán:

= (6.25 ×  6.023 * 10²³) moléculas de oxígeno.

≅ 3.764 × 10²³ moléculas de oxígeno.

(c) El número de átomos de oxígeno es:

= 2 × 3.764 × 10²³

= 7.528 × 10²³ átomos de oxígeno

(d) Usando la ecuación de gas ideal

PV = nRT

El volumen ocupado por el oxígeno = \dfrac{nRT}{P}

Volumen ocupado por oxígeno = \dfrac{ 6.25 * 8.314 *273}{1}

Volumen ocupado por oxígeno= 14185.76 m³

3 0
3 years ago
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