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3 years ago

Determina la masa molar y el volumen que ocupa la siguiente sustancia CO2, si su masa es de 28 gr. *

Mathematics

1 answer:

balandron [24]3 years ago

7 0

Answer:

Para el CO₂ sabemos que:

densidad = 0,001976 g/cm³

Sabemos que:

densidad = masa/volumen

Entonces, si tenemos una masa de 28 g, podemos escribir:

volumen = masa/densidad

volumen = (28g)/(0,001976 g/cm³) = 14,170 cm^3

Para obtener la masa molar (es decir, la masa de un mol de esta sustancia) simplemente sumamos la masa de un mol de cada componente.

Carbono: tiene una masa molar de 12 g/mol

Oxígeno: tiene una masa molar de 16 g/mol (y tenemos dos oxígenos)

entonces la masa molar va a ser:

masa molar = 12g/mol + 2*16g/mol = 44 g/mol

Es decir, un mol de CO₂, pesa 44 gramos.

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Match the hyperbolas represented by the equations to their foci.

Arte-miy333 [17]

Answer:

1) (1 , -22) and (1 , 12) ⇔ (y + 5)²/15² - (x - 1)²/8² = 1

2) (-7 , 5) and (3 , 5) ⇔ (x + 2)²/3² - (y - 5)²/4² = 1

3) (-6 , -2) and (14 , -2) ⇔ (x - 4)²/8² - (y + 2)²/6² = 1

4) (-7 , -10) and (-7 , 16) ⇔ (y - 3)²/5² - (x + 7)²/12² = 1

Step-by-step explanation:

* Lets study the equation of the hyperbola

- The standard form of the equation of a hyperbola with

center (h , k) and transverse axis parallel to the x-axis is

(x - h)²/a² - (y - k)²/b² = 1

- the coordinates of the foci are (h ± c , k), where c² = a² + b²

- The standard form of the equation of a hyperbola with

center (h , k) and transverse axis parallel to the y-axis is

(y - k)²/a² - (x - h)²/b² = 1

- the coordinates of the foci are (h , k ± c), where c² = a² + b²

* Lets look to the problem

1) The foci are (1 , -22) and (1 , 12)

- Compare the point with the previous rules

∵ h = 1 and k ± c = -22 ,12

∴ The form of the equation will be (y - k)²/a² - (x - h)²/b² = 1

∵ k + c = -22 ⇒ (1)

∵ k - c = 12 ⇒ (2)

* Add (1) and(2)

∴ 2k = -10 ⇒ ÷2

∴ k = -5

* substitute the value of k in (1)

∴ -5 + c = -22 ⇒ add 5 to both sides

∴ c = -17

∴ c² = (-17)² = 289

∵ c² = a² + b²

∴ a² + b² = 289

* Now lets check which answer has (h , k) = (1 , -5)

and a² + b² = 289 in the form (y - k)²/a² - (x - h)²/b² = 1

∵ 15² + 8² = 289

∵ (h , k) = (1 , -5)

∴ The answer is (y + 5)²/15² - (x - 1)²/8² = 1

* (1 , -22) and (1 , 12) ⇔ (y + 5)²/15² - (x - 1)²/8² = 1

2) The foci are (-7 , 5) and (3 , 5)

- Compare the point with the previous rules

∵ k = 5 and h ± c = -7 ,3

∴ The form of the equation will be (x - h)²/a² - (y - k)²/b² = 1

∵ h + c = -7 ⇒ (1)

∵ h - c = 3 ⇒ (2)

* Add (1) and(2)

∴ 2h = -4 ⇒ ÷2

∴ h = -2

* substitute the value of h in (1)

∴ -2 + c = -7 ⇒ add 2 to both sides

∴ c = -5

∴ c² = (-5)² = 25

∵ c² = a² + b²

∴ a² + b² = 25

* Now lets check which answer has (h , k) = (-2 , 5)

and a² + b² = 25 in the form (x - h)²/a² - (y - k)²/b² = 1

∵ 3² + 4² = 25

∵ (h , k) = (-2 , 5)

∴ The answer is (x + 2)²/3² - (y - 5)²/4² = 1

* (-7 , 5) and (3 , 5) ⇔ (x + 2)²/3² - (y - 5)²/4² = 1

3) The foci are (-6 , -2) and (14 , -2)

- Compare the point with the previous rules

∵ k = -2 and h ± c = -6 ,14

∴ The form of the equation will be (x - h)²/a² - (y - k)²/b² = 1

∵ h + c = -6 ⇒ (1)

∵ h - c = 14 ⇒ (2)

* Add (1) and(2)

∴ 2h = 8 ⇒ ÷2

∴ h = 4

* substitute the value of h in (1)

∴ 4 + c = -6 ⇒ subtract 4 from both sides

∴ c = -10

∴ c² = (-10)² = 100

∵ c² = a² + b²

∴ a² + b² = 100

* Now lets check which answer has (h , k) = (4 , -2)

and a² + b² = 100 in the form (x - h)²/a² - (y - k)²/b² = 1

∵ 8² + 6² = 100

∵ (h , k) = (4 , -2)

∴ The answer is (x - 4)²/8² - (y + 2)²/6² = 1

* (-6 , -2) and (14 , -2) ⇔ (x - 4)²/8² - (y + 2)²/6² = 1

4) The foci are (-7 , -10) and (-7 , 16)

- Compare the point with the previous rules

∵ h = -7 and k ± c = -10 , 16

∴ The form of the equation will be (y - k)²/a² - (x - h)²/b² = 1

∵ k + c = -10 ⇒ (1)

∵ k - c = 16 ⇒ (2)

* Add (1) and(2)

∴ 2k = 6 ⇒ ÷2

∴ k = 3

* substitute the value of k in (1)

∴ 3 + c = -10 ⇒ subtract 3 from both sides

∴ c = -13

∴ c² = (-13)² = 169

∵ c² = a² + b²

∴ a² + b² = 169

* Now lets check which answer has (h , k) = (-7 , 3)

and a² + b² = 169 in the form (y - k)²/a² - (x - h)²/b² = 1

∵ 5² + 12² = 169

∵ (h , k) = (-7 , 3)

∴ The answer is (y - 3)²/5² - (x + 7)²/12² = 1

* (-7 , -10) and (-7 , 16) ⇔ (y - 3)²/5² - (x + 7)²/12² = 1

7 0

3 years ago

if a distribution of raw scores were plotted and then the scores were transformed to z scores, would the shape of the distributi

nadya68 [22]

<span>•<span>If the raw score is transformed into a z-score, however, the value of the z-score tells exactly where the score is located relative to all the other scores in the distribution. </span></span>

3 0

3 years ago

Solve for n.<br><br> 6 + 2n = 10<br><br> n =

melisa1 [442]

Step-by-step explanation:

<h2>Answer:-</h2>

$\tt{6+2n = 10}$

$\tt{2n = 10-6}$

$\tt{2n = 4}$

$\tt{n = \dfrac{4}{2}}$

$\boxed{\tt{n=2}}$

Hope it helps :)

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3 years ago

There are 3 consecutive integers. Four times the smallest integer is equal to six more than three times the largest integer. Wha

Aliun [14]

Answer:

Step-by-step explanation:

Let the smallest integer equal x - 1, the next one equal x, and the last one equal x + 1.

4(x-1) = 3(x-1) + 6

x-1 = 5

x = 6

The smallest integer is x - 1, which is equal to 5. Thus, the smallest of the three integers is 5.

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2 years ago

Jeremy is taking a photography class, but he doesn't own a camera. The class organizers will rent him a camera for $6 per day. J

Mazyrski [523]

Solving the given inequality for d, we get...

6d + 15 < 50
6d + 15-15 < 50-15 <<-- subtract 15 from both sides
6d < 35
6d/6 < 35/6 <<--- divide both sides by 6
d < 5.83

Which means that d can be any of the values in this set: {0, 1, 2, 3, 4, 5}

The smallest d can be is 0. In this scenario, Jeremy pays the $15 registration but doesn't rent the camera at all
The largest d can be is 5. In this scenario, Jeremy rents the camera for 5 days
Any larger value of d is not allowed as it would make the total cost go over $50
Notice how I'm rounding down regardless how close 5.83 is to 6

8 0

3 years ago

Read 2 more answers