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Kruka [31]
3 years ago
6

What is located within 10 to 15 km above the earth's surface?​

Chemistry
1 answer:
mihalych1998 [28]3 years ago
8 0

It extends 10 to 15 kilometers (6 to 10 miles) above the earth's surface depending on latitude and time of the year. The troposphere contains most of the atmosphere's water vapor.

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Chloroform has a density of 1.483g/ml. A sample of chloroform has a volume of 1.93 liters. How many kilograms does it weight? ​
ollegr [7]

Weight of Chloroform : = 2.862 kg

<h3>Further explanation</h3>

Given

Density 1.483 g/ml

Volume = 1.93 L

Required

Weight of Chloroform

Solution

Density is a quantity derived from the mass and volume  

Density is the ratio of mass per unit volume  

Density formula:  

\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}

ρ = density  

m = mass  

v = volume  

Convert density to kg/L :

=1.483g/ml = 1.483 kg/L

So the weight(mass) :

= ρ x V

= 1.483 kg/L x 1.93 L

= 2.862 kg

8 0
3 years ago
A 75 g piece of gold (Au) at 1000 K is dropped into 200 g of H2O at 300K in an insulatedcontainer at 1 bar. Calculate the temper
Arturiano [62]

Answer:

the final temperature is T final = 308 K

Explanation:

since all heat released by gold is absorbed by water

Q gold + Q water = Q surroundings =0 (insulated)

Assuming first that no evaporation of water occurs , and denoting g as gold and w as water , then

Q gold = m g*cp g* ( T final - T initial g)

Q gold = m w*cp w* ( T final - T initial w)

where

m= mass

cp = specific heat capacity

T final = final temperature

T initial g and T initial w =  initial temperature of gold and water respectively

thus

Q gold + Q water = 0

m g*cp g* ( T final - T initial g) + m w*cp w* ( T final - T initial w) =0

m g*cp g* T final + m w*cp w* T final =  m g*cp g* T initial g+ m w*cp w* T initial w

T final = (m g*cp g* T initial g+ m w*cp w* T initial w)/(m g*cp g+ m w*cp w)

replacing values and assuming cp w = 1 cal/gK = 4.816 J/gK and cp g = 0.129 J/gK (from tables), then

T final =  (75 g*0.129 J/gK* 1000 K + 200 g * 4.816 J/gK * 300 K )/(75 g*0.129 J/gK*+ 200 g * 4.816 J/gK ) = 308 K

T final = 308 K

since T boiling water = 373 K and T final = 308 K , we confirm that water does not evaporate

therefore the final temperature is T final = 308 K

3 0
3 years ago
What is activation energy
Stolb23 [73]
The minimum amount of energy that colliding particles must have for them to react.
4 0
3 years ago
Read 2 more answers
How many grams of potassium sulphate are formed when 49g of sulphuric acid in solution are neutralised by potassium hydroxide?
Katena32 [7]
C - keep counting to the atoms
6 0
2 years ago
For the reaction: CH3NH2(aq) + H2O(aq) ⇌ CH3NH3 +(aq) + OH- Determine the change in the pH (ΔpH) for the addition of 6.7 M CH3NH
Korolek [52]

Answer:

The change in the pH (ΔpH) is 2,17

Explanation:

The reaction:

CH₃NH₂(aq) + H₂O(aq) ⇌ CH₃NH₃⁺(aq) + OH⁻

kb = \frac{[OH^{-}][CH_{3}NH_{3}^+]}{[CH_{3}NH_{2}]} <em>(1)</em>

In equilibrium, a solution of CH₃NH₂ 4,7M produces:

[CH₃NH₂] = 4,7 - x

[CH₃NH₃⁺] = x

[OH⁻] = x

Replacing in (1):

4,38x10^{-4} = \frac{x^2}{4,7-x}

x² + 4,38x10⁻⁴x - 2,0586x10⁻³ = 0

The solutions are:

x = -0,0456 No physical sense. There are not negative concentrations.

x = 0,04515 Real answer.

The concentration of [OH⁻] is 0,04515 M.

As pOH = -log [OH⁻] And pH+pOH = 14. The pH of this solution is:

<em>pH = 12,65</em>

The addition of 6,7M produce this changes in concentrations:

[CH₃NH₂] = 4,656 + x

[CH₃NH₃⁺] = 6,74515 - x

[OH⁻] = 0,04515 - x

Replacing in (1) you will obtain:

x² - 6,7907x + 0,3025 = 0

Solving for x:

x = 6,74586 No physical sense

x = 0,04484 Real answer.

Thus, [OH⁻] = 0,04515 - 0,044842 = 3,08x10⁻⁴M

pOH = 3,51.

<em>pH = 10,49</em>

Thus ΔpH is 12,65 - 10,49 = <em>2,16 ≈ 2,17</em>

I hope it helps!

4 0
3 years ago
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