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3 years ago

What do Hubris mean?

Chemistry

2 answers:

Alona [7]3 years ago

7 0

Hubris: excessive pride or self-confidence.

Margarita [4]3 years ago

5 0

I believe the answer is mostly pride

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8. The time period of artificial satellite in a circular orbit of radius R is T. The radius of the orbit in which time period is

Elena-2011 [213]

Explanation:

It is given that,

The time period of artificial satellite in a circular orbit of radius R is T. The relation between the time period and the radius is given by :

$T^2\propto R^3$

The radius of the orbit in which time period is 8T is R'. So, the relation is given by :

$(\dfrac{T}{T'})^2=(\dfrac{R}{R'})^3$

$(\dfrac{T}{8T})^2=(\dfrac{R}{R'})^3$

$\dfrac{1}{64}=(\dfrac{R}{R'})^3$

$R'=4\times R$

So, the radius of the orbit in which time period is 8T is 4R. Hence, this is the required solution.

4 0

3 years ago

Cosmic rays are

IrinaK [193]

A. High energy radiation produced in the ozone layer. (:

6 0

2 years ago

Cuántos moles hay en 17.5 gramod de ZnCl2

Aliun [14]

Explanation:

Zn=65

Cl2= 35+35=70

65+70=135g

1 mole ZnCl2 = 135g

x mole = 17.5g

17.5g × 1 mole/ 135g= 0.129 moles en 17.5g de ZnCl2

4 0

3 years ago

A solution is prepared by mixing 93.0 mL of 5.00 M HCl and 37.0 mL of 8.00 M HNO3. Water is then added until the final volume is

Charra [1.4K]

Answer:

$[H^{+}] = 0.761 \frac{mol}{L}$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

$pH = 0.119$

Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

$n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L) \\ n_{H^{+} } from HNO_{3} = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)$

Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows

$n_{H^{+} } from HCl = (5.00)(0.093)$

$n_{H^{+} } from HCl = 0.465 mol$

$n_{H^{+} } from HNO_{3} = (8.00)(0.037)$

$n_{H^{+} } from HNO_{3} = 0.296 mol$

$n_{H^{+}(total) } = 0.296 + 0.465$

$n_{H^{+}(total) } = 0.761 mol$

For molar concentration of hydrogen ions:

$[H^{+}] = \frac{n_{H^{+}}(mol)}{V(L)}$

$[H^{+}] = \frac{0.761}{1.00}$

$[H^{+}] = 0.761 \frac{mol}{L}$

From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows

$K_{w} = [H^{+} ][OH^{-} ]$

$[OH^{-}]=\frac{Kw}{[H^{+}] }$

$[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

The pH of the solution can be measured by the following formula:

$pH = -log[H^{+} ]$

$pH = -log(0.761)$

$pH = 0.119$

5 0

3 years ago

PLEASE HELP! Class is almost ending, I don't want to repeat.

nikklg [1K]

ANSWERS:

1. $2KClO_{3}$ ⇒ 2KCl + $3O_{2}$

2. $H_{2} C_{2} O_{4} + 2NaOH$ ⇒ $Na_{2} C_{2} O_{4} + 2 H_{2} O$

3. $2 C_{10} H_{22} + 31O_{2}$ ⇒ $20C O_{2} + 22 H_{2} O$

4. $4 H_{2} + Fe_{3} O_{4}$ ⇒ $3Fe +4 H_{2} O$

5. $4P + 5 O_{2}$ ⇒ $2 P_{2} O_{5}$

I have attached my work for 1-3. I would like to see if you can get 4 and 5 on your own. but, if you are struggling/copnfused please let me know in the comments! :)

6 0

3 years ago