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AlekseyPX
3 years ago
13

A student combined equal amounts of two solutions. One solution had a pH of 2 and the other had a pH of 12. Which would most lik

ely be the resulting pH?
A. 1
B. 3
C. 6
D.11
Physics
2 answers:
zalisa [80]3 years ago
6 0
Answer: optio C. 6.

pH measures the acidity of the solutions.

pH 2 is highly acid and pH 12 is highly basic (alkalyne).

You can be sure that the pH of the solution will be intermediate between 2 and 12.

Given that the volumen of the two solutions are equal the pH of the resulting combination cannot be so close to the original solutions. That takes you to eliminate all the options except the option C. 6.

You cannot be sure that the final pH will be 6, because that will depend on the stregths of the two original solutions, that is why the question asks for the most likely resulting pH, and the answer is optio C. 6. 
baherus [9]3 years ago
4 0

Answer:

C

Explanation:

the answer is C

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amm1812
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Summary:

-- Equations #2, #3, #4, and #6 are all correct statements,
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7 0
3 years ago
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The mass of a hot-air balloon and its cargo (not including the air inside) is 170 kg. The air outside is at 10.0°C and 101 kPa.
scoundrel [369]

Answer:

108.37°C

Explanation:

P₁ = Initial pressure = 101 kPa

V₁ = Initial volume = 530 m³

T₁ = Initial temperature = 10°C = 10+273.15 =283.15 K

P₂ = Final pressure = 101 kPa (because it is open to atmosphere)

V₂ = Final volume = 530 m³

P₁V₁ = n₁RT₁

⇒101×530 = n₁RT₁

⇒53530 J = n₁RT₁

P₂V₂ = n₂RT₂

⇒53530 J = n₂RT₂

\frac{m_1}{m_2}=\frac{\rho V_1}{\rho V_1-170}\\\Rightarrow \frac{m_1}{m_2}=\frac{1.244\times 530}{1.244\times 530-170}=1.347\\\Rightarrow \frac{m_1}{m_2}=1.347\\\Rightarrow \frac{n_1}{n_2}=1.347

Dividing the first two equations we get

1=\frac{n_1}{n_2}\frac{T_1}{T_2}\\\Rightarrow 1=1.347\frac{283.15}{T_2}\\\Rightarrow T_2=1.347\times 283.15= 381.52\ K

∴Temperature must the air in the balloon be warmed before the balloon will lift off is 381.25-273.15 = 108.37°C

8 0
3 years ago
A satellite of Mars, called Phobos, has an orbital radius of 9.4 ✕ 106 m and a period of 2.8 ✕ 104 s. Assuming the orbit is circ
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Answer:

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Explanation:

assume

M= mass of Mars

m=mass of phobos

r=orbital radius

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we can apply F=ma to this orbital motion (considering the cricular motion laws)

where,

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