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klio [65]
3 years ago
12

How would you know if this combination is likely to be found in dirt? Please explain!

Chemistry
2 answers:
Natasha2012 [34]3 years ago
5 0
No at least I don’t think there would be
Elan Coil [88]3 years ago
4 0
No it is not likely. That is a ratio of 10:4 N^14 and N^15 which doesn’t work. It needs a higher amount
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Two solid samples each contain sulfur, oxygen, and sodium, only. These samples have the same color, melting point, density, and
katovenus [111]
<span>The correct answer is 1. compound. There are three elements and not just one so it's not element. Mixtures wouldn't have the same properties that you described but these do so it's a compounds. Solution is not a solid thing but rather a liquid one. The two samples are therefore compounds.</span>
7 0
3 years ago
When a solid undergoes _____ it takes up less space due to cooling
almond37 [142]

Answer:

solificATION

Explanation:

4 0
3 years ago
Read 2 more answers
The trait for flower color in a plant has red and white alleles. The red color is the dominant trait. What is the phenotypic rat
Fudgin [204]
Answer :

2 red : 2 white

Explanation:

;)long story short

hetro with hetro gives 3:1
hetro with recessive gives 1:1

Hetro :- ( Rr) one capital letter and one small
Recessive :- (rr) two small leters

6 0
3 years ago
2.00 L of a gas is collected at 25.0°C and 745.0 mmHg. What is the volume at 760.0 mmHg
Leviafan [203]

1.7960L

Explanation:

the mass of the gas is constant in both instances

pv/T=constant(according to pv=nRT)

745mmHg*2L/298K=760mmHg*v/273K

v=1.7960L

5 0
3 years ago
Determine el PH y el % de disociación de una solución de ácido débil, sabiendo que se disuelven 20 gramos del ácido (masa molar=
IceJOKER [234]

Answer:

pH = 4.27. Porcentaje de disociación: 0.03%

Explanation:

El pH de un ácido débil, HX, se obtiene haciendo uso de su equilibrio:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

Donde la constante de equilibrio, Ka, es

Ka = 1.65x10⁻⁸ = [H⁺] [X⁻] / [HX]

Como los iones H⁺ y X⁻ vienen del mismo equilibrio podemos decir:

[H⁺] = [X⁻]

[HX] es:

20g * (1mol/55g) = 0.3636moles / 2.100L = 0.1732M

Reemplazando es Ka:

1.65x10⁻⁸ = [H⁺] [H⁺] / [0.1732M]

2.858x10⁻⁹ = [H⁺]²

5.35x10⁻⁵M = [H⁺]

pH = -log[H⁺]

<h3>pH = 4.27</h3>

El porcentaje de disociacion es [X⁻] / [HX] inicial * 100

Reemplazando

5.35x10⁻⁵M / 0.1732M * 100

<h3>0.03%</h3>
5 0
3 years ago
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