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umka21 [38]
3 years ago
15

Ejemplos de magnitudes escalares en la vida cotidiana?​

Chemistry
1 answer:
Doss [256]3 years ago
3 0

Answer:

Temperatura. Es una magnitud escalar ya que un valor numérico la define por completo. ...

Presión. ...

Longitud. ...

Energía. ...

Masa. ...

Tiempo. ...

Área. ...

Volumen.

Explanation:

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C5H12 + 8O2 → 5CO2 + 6H2O<br><br> Calculate the energy change for the reaction:
KATRIN_1 [288]
C5H12 (l) + 8O2 (g) ----> 5CO2 (g) + 6H2O (l)
Delta H = -3505.8 kJ/mol

C (s) + O2 (g) -----> CO2 (g)
Delta H = -393.5 kJ/mol

H2 (g) + (1/2)O2 (g) ------> H2O (l)
Delta H = -286 kJ/mol

Possible answers:
a. +35 kJ/mol
b. + 1,073 kJ/mol
c. -4,185 kJ/mol
d. -2,826 kJ/mol
e. -178 kJ/mol
8 0
3 years ago
Helppppp with this pleasee
vazorg [7]

Answer:

A. 1, 2, 5

Explanation:

Count the number of Ns in the formula.

- Hope that helped! Please let me know if you need a further explanation.

3 0
3 years ago
According to the theory of plate tectonics,
11Alexandr11 [23.1K]
A I’m sure of it because it only makes since one would think ✊
3 0
3 years ago
I need help please! I will mark brainliest
Fittoniya [83]

Answer: I think it's C

Explanation: I hope this helps (Sorry if it doesn't)

8 0
1 year ago
Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1
givi [52]

According to Hasselbach-Henderson equation:

pH=pK_{a}+log\frac{[A^{-}]}{[HA]}

Here, [A^{-}] is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is CH_{3}COOK and acid is CH_{3}COOH thus, Hasselbach-Henderson equation will be as follows:

pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{2}{1}=5.06

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{1}{3}=4.28

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{5}{1}=5.45

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{1}{1}=4.76

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{1}{10}=3.76

Therefore, pH of solution is 3.76.

4 0
3 years ago
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