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4 years ago

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What is earths spheres

2 answers:

Oliga [24]4 years ago

6 0

Geosphere >> Which is Earth's entire solid body..

Hydrosphere >> Which is all the "liquid" water on Earth..

Atmosphere >> Which is the few layers of "gasses" that are surrounding Earth..

Cryosphere >> Which is all the "frozen" water on Earth..

Biosphere >> Us! All the living organisms on Earth..

Anton [14]4 years ago

6 0

<span>Everything in Earth's system can be placed into one of four major subsystems: land, water, living things, or air. These four subsystems are called "spheres." Specifically, they are the "lithosphere" land, "hydrosphere" water, "biosphere" living things, and "atmosphere" air. hope this helped </span>

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A tablet of Pain Be Gone Aspirin, which had a mass of 1.213 g, was pulverized and 1.159 g were dissolved in 10.0 mL of ethyl alc

allochka39001 [22]

Answer:

Moles of NaOH that reacted: $n_{NaOH}=1.64*10^{-3}mol$

Explanation:

During the titration, all moles of NaOH added in the solution, react with the acetylsalicylic acid neutralizing each other.

The titration required 15.62 ml (0.01562 L) of the 0.1052 M NaOH solution. The moles of NaOH in that volume are:

$n_{NaOH}=0.01562 L * \frac{0.1052mol}{L}$

$n_{NaOH}=1.64*10^{-3}mol$

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3 years ago

What are some ways we can improve the chemical reactions in batteries to increase their efficiency?

Ainat [17]

Answer:

Charging Level The cycle life of Lithium batteries can be increased by reducing the charging cut off voltage.

Explanation:

6 0

3 years ago

A photon of light possesses 5 x 10^-19 J of energy. Calculate its frequency

saveliy_v [14]

Answer:

The frequency of photon is 0.75×10¹⁵ s⁻¹.

Explanation:

Given data:

Energy of photon = 5×10⁻¹⁹ J

Frequency of photon = ?

Solution:

Formula;

E = hf

h = planck's constant = 6.63×10⁻³⁴ Js

5×10⁻¹⁹ J = 6.63×10⁻³⁴ Js ×f

f = 5×10⁻¹⁹ J / 6.63×10⁻³⁴ Js

f = 0.75×10¹⁵ s⁻¹

The frequency of photon is 0.75×10¹⁵ s⁻¹.

4 0

3 years ago

At a certain temperature, 0.3411 0.3411 mol of N 2 N2 and 1.661 1.661 mol of H 2 H2 are placed in a 2.50 2.50 L container. N 2 (

Sveta_85 [38]

<u>Answer:</u> The equilibrium constant for the above reaction is 1.31

<u>Explanation:</u>

We are given:

Initial moles of nitrogen gas = 0.3411 moles

Initial moles of hydrogen gas = 1.661 moles

Equilibrium moles of nitrogen gas = 0.2001 moles

For the given chemical reaction:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

<u>Initial:</u> 0.3411 1.661

<u>At eqllm:</u> 0.3411-x 1.661-3x 2x

Evaluating for 'x', we get:

$\Rightarrow (0.3411-x)=0.2001\\\\\Rightarrow x=0.3411-0.2001=0.141$

Volume of the container = 2.50 L

The expression of $K_c$ for the above equation follows:

$K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}$

We are given:

$[NH_3]=\frac{2\times 0.141}{2.50}=0.1128M$

$[N_2]=\frac{0.2001}{2.5}=0.08004M$

$[H_2]=\frac{1.661-(3\times 0.141)}{2.5}=0.4952M$

Putting values in above expression, we get:

$K_c=\frac{(0.1128)^2}{0.08004\times (0.4952)^3}\\\\K_c=1.31$

Hence, the equilibrium constant for the above reaction is 1.31

5 0

3 years ago

A 225. mL sample 3.5M solution of KBr is diluted to a final volume of 750. mL. What is the final concentration of this solution?

Allushta [10]

C= comcentration
V= volume
C1*V1=C2*V2
3.5M*225.mL = xM*750 mL
x=(3.5*225/750) =1.05M≈1.1M

7 0

4 years ago