DIRECTION: Supply the missing information about the scientist listed in the graphic
organizer.

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Answer:

biosphere and lithosphere

Explanation:

The biosphere is described as the zone of life on Earth. It is a sum of all ecosystems. The biosphere is composed of living organisms and non-living factors.

The lithosphere is the outer part of the Earth such that this part is rocky. The lithosphere is made up of the brittle crust.

In general, chemicals enter Ecosystems through the biosphere and lithosphere.

Answer:

22 g

Explanation:

Step 1: Write the balanced equation

N₂(g) + 3 H₂(g) ⇒ 2 NH₃(g)

Step 2: Calculate the moles of NH₃ produced from 2.0 moles of H₂

The molar ratio of H₂ to NH₃ is 3:2.

2.0 mol H₂ × 2 mol NH₃/3 mol H₂ = 1.3 mol NH₃

Step 3: Calculate the mass corresponding to 1.3 moles of NH₃

The molar mass of NH₃ is 17.03 g/mol.

1.3 mol × 17.03 g/mol = 22 g

Answer:

[tex]V=5.2 mL=0.052L[/tex]

Explanation:

Hello!

In this case, since the chemical reaction between silver nitrate and sodium chloride is:

[tex]AgNO_3(aq)+NaCl(aq)\rightarrow AgCl(s)+NaNO_3(aq)[/tex]

We can see there is a 1:1 mole ratio between each solution; thus, we first compute the moles of each reactant considering their molar masses:

[tex]n_{AgNO_3}=14.77g*\frac{1mol}{169.87g}=0.087molAgNO_3\\\\ n_{NaCl}=0.2631g*\frac{1mol}{58.44}=0.0045molNaCl[/tex]

Now, since the concentration of the silver chloride solution is 0.087 M, we may assume that the concentration of the NaCl solution is the same, so we can compute the volume as shown below:

[tex]V=\frac{n_{NaCl}}{M}=\frac{0.0045mol}{0.087mol/L}\\\\V=0.052L[/tex]

Or:

[tex]V=5.2 mL[/tex]

Best regards!

The volume of solution needed to react with 0.2631 g of NaCl  is 0.052 L.

Volume of the solution will be calculated by using the below formula:

M = n/V, where

M = concentration in terms of molarity

n = no. of moles

V = volume

Given chemical reaction is:

AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

First we calculate the moles of given reactants by using the formula:

n = W/M , where

W = given mass

M = molar mass

Moles of AgNO₃ = 14.77g / 169.87g/mole = 0.087 mole

Moles of NaCl = 0.2631g / 58.44g/mole = 0.0045 mole

Concentration of AgNO₃ = 0.087 mole / 1L = 0.087M

From the stoichiometry of the reaction it is clear that mole ration of AgNO₃ & NaCl is 1:1. So, we take the concentration of NaCl is equal to the concentration of AgNO₃ and calculate the volume by using the above formula as:

Volume of NaCl = 0.0045mole / 0.087M = 0.052 L

Hence, 0.052 L is the required volume of NaCl.

To know more about moles, visit the below link:

https://brainly.com/question/17199947

Answer:

the steady-state concentration of TCE in the treated water leaving the reactor is 5.88 mg/L  

Explanation:

Given that;

Tank volume v = 3250 liters

wastewater flows into the tank Q = 200 L/min

TCE concentration Co= 25 mg/L

reactor decays TCE at a reaction rate K = 0.20 min-1

mass balance

we know that;

Accumulation = inflow - outflow ± generation

⇒dc/dt = QCo - Qc ± rc.V

now at a steady state; dc/dt = 0

so

0 = QCo - Qc + rcV

where rc = -kc

0 = QCo - Qc - kcV

Qc + kcV = QCo

c(Q + kV) = QCo  

c = QCo / (Q + kV)

so we substitute

c = (200 × 25) / (200 + (0.2×3250))

c = 5000 / 850

c = 5.88 mg/L      

Therefore,  the steady-state concentration of TCE in the treated water leaving the reactor is 5.88 mg/L  

Answer: Thus 24.0 g of [tex]SO_2[/tex] would be needed.

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]      

[tex]\text{Moles of} O_2=\frac{6.00g}{32g/mol}=0.1875moles[/tex]

[tex]2SO_2(g)+O_2(g)\rightarrow 2SO_3(l)[/tex]  

According to stoichiometry :

1 mole of [tex]O_2[/tex] require = 2 moles of [tex]SO_2[/tex]

Thus 0.1875 moles of [tex]O_2[/tex] will require=[tex]\frac{2}{1}\times 0.1875=0.375moles[/tex]  of [tex]SO_2[/tex]  

Mass of [tex]SO_2=moles\times {\text {Molar mass}}=0.375moles\times 64g/mol=24.0g[/tex]

Thus 24.0 g of [tex]SO_2[/tex] would be needed to completely react with 6.00 g of [tex]O_2[/tex] such that all reactants could be consumed.

Answer:

[tex]m_{Hg}=829.4gHg[/tex]

Explanation:

Hello there!

In this case, considering the Avogadro's number, which helps us to realize that 1 mole of mercury atoms contains 6.022x10²³ atoms and at the same time 1 mole of mercury weights 200.59 g, we obtain:

[tex]m_{Hg}=2.49x10^{24}atoms*\frac{1mol}{6.022x10^{23}atoms} *\frac{200.59g}{1molHg}\\\\m_{Hg}=829.4gHg[/tex]

Best regards!

Answer:

If your asking what golds natural state of matter is it's solid.

Explanation:

Answer:

the answer is soild

Explanation:

i did it on edge :)