answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
vlada-n [284]
2 years ago
6

How might the biosphere Be effected if a harmful sub stance Entered the Geosphere

Chemistry
1 answer:
Marina86 [1]2 years ago
6 0

Explanation:

The biosphere will be affected if harmful substances enters into the geosphere. In fact, natural contaminants from the geosphere are known to impose serious health and environmental hazards.

The biosphere is the portion of earth where life is constrained.

The geosphere is the solid portion of the earth.

  • The geosphere and biosphere continuously interacts.
  • For example, the soil we cultivate our food is a product of the geosphere.

If for example,  a poisonous element like lead abounds geologically in a places.

When surface water or underground water passes through, the lead is becomes dissolved in them. When life on earth depends on water. On using the lead laden water, harmful lead poisoning occurs.

This is one of the most typical way in which the geosphere affects the biosphere.

learn more:

Food contaminants brainly.com/question/4646230

#learnwithBrainly

You might be interested in
Which of the following solutions is a buffer?A) A solution made by mixing 100 mL of 0.100 M HClO and 50 mL of 0.100 M HCl.B) A s
kumpel [21]

Answer:

D

Explanation:

solution made by mixing 100 mL of 0.100 M HClO and 50 mL of 0.100 M NaOH Can resist pH change when there is little addition of either acid or base, hence it is a buffer solution

7 0
2 years ago
If 200.0 g of copper(II) sulfate react with an excess of zinc metal, what is the theoretical yield of copper? 1.253 g 50.72 g 79
Helen [10]
1)we need a balanced equation: CuSO₄ + Zn ---> ZnSO₄ + Cu

2) we need to convert the grams of CuSO₄ to moles using the molar mass. 

molar mass CuSO₄= 63.5 + 32.0 + (4 x 16.0)= 160 g/mol

200.0 g CuSO_4 ( \frac{1 mol}{160 grams} )= 1.25 mol CuSO_4

3) convert moles of CuSO₄ to moles of Cu

1.25 mol CuSO_4 ( \frac{1 mol Cu}{1 mol CuSO_4} )= 1.25 mol Cu

4) convert moles of Cu to grams using it's molar mass.

molar mass Cu= 63.5 g/mol

1.25 mol (\frac{63.5 grams}{1 mol} )= 79.4 grams Cu

I did it step-by-step as the explanation but you can do all of this in one step. 

200.0 g CuSO_4 ( \frac{1 mol CuSO_4}{160 g} ) ( \frac{1 mol Cu}{1 mol CuSO_4} ) ( \frac{63.5 grams}{1 mol Cu} )= 79.4 grams Cu


4 0
2 years ago
10 points) Given the following two half-reactions, write the overall reaction in the direction in which it is product-favored, a
IceJOKER [234]

Answer:

The over all reaction :

2Fe^{3+}(aq)+Pb(s)\rightarrow 2Fe^{2+}(s)+Pb^{2+}(aq)  

The standard cell potential of the reaction is 0,.897 Volts.

Explanation:

Reduction at cathode :

Fe^{3+}(aq)+e^-\rightarrow Fe^{2+}(s)..[1]  

E^o_{Fe^{3+}/Fe^{2+}}=0.771 V

Reduction potential of Fe^{3+} to Fe^{2+}=0.771 V

Oxidation at anode:

Pb(s)\rightarrow Pb^{2+}(aq)+2e^-.[2]

E^o_{Pb^{2+}/Pb}=-0.126 V

Reduction potential of Pb^{3+} to Pb=-0.126 V

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}

Putting values in above equation, we get:

E^o_{cell}=E^o_{Fe^{3+}/Fe^{2+}}-E^o_{Pb^{2+}/Pb}

=0.771 V-(-0.126 )=0.897 V

The over all reaction : 2 × [1] + [2]

2Fe^{3+}(aq)+Pb(s)\rightarrow 2Fe^{2+}(s)+Pb^{2+}(aq)  

The standard cell potential of the reaction is 0,.897 Volts.

7 0
2 years ago
What is the concentration of a solution made with 0.150 moles of KOH in 400.0 mL of solution?
otez555 [7]

Answer:

concentration = \frac{0.15}{0.4}=0.375 mol/L

Explanation:

Concentration: i is defined as the mole per litre.

concentration = \frac{mole}{volume in L}

mole=0.15

volume=400 ml=0.4 litre

concentration = \frac{0.15}{0.4}=0.375 mol/L

6 0
2 years ago
What is the percent yield of this reaction if 22.o g of Mgl2 is produced by the reaction of 25.0 g of Mg with 25.0 g of l2?
expeople1 [14]

% yield = 80.719

<h3>Further explanation</h3>

Given

22.0 g of Mgl₂

25.0 g of Mg

25.0 g of l₂

Required

The percent yield

Solution

Reaction

Mg + I₂⇒ MgI₂

mol Mg = 25 g : 24.305 g/mol = 1.029

mol I₂ = 25 g : 253.809 g/mol = 0.098

Limiting reactant = I₂

Excess reactant = Mg

mol MgI₂ based on I₂, so mol MgI₂ = 0.098

Mass MgI₂ (theoretical):

= mol x MW

= 0.098 x 278.114

= 27.255 g

% yield = (actual/theoretical) x 100%

% yield = (22 / 27.255) x 100%

% yield = 80.719

8 0
1 year ago
Other questions:
  • PLEASE HELP!!! Which of John Dalton's contributions are present in the modern atomic model and which were eventually disproven a
    12·1 answer
  • A 226.4-l cylinder contains 65.5% he(g) and 34.5% kr(g) by mass at 27.0°c and 1.40 atm total pressure. what is the mass of he in
    13·2 answers
  • What is the poh of an aqueous solution at 25.0 °c that contains 1.35 × 10-8 m hydroxide ion?
    9·1 answer
  • An air mass of volume 6.5 x 10 to the fifth L starts at sea level, where the pressure is 775 mm HG. It rises up a mountain where
    12·1 answer
  • You run an electrolysis of NaCl and collect one of the products in a test tube. You realize later you did not label the test tub
    10·2 answers
  • You are performing a simple distillation of roughly 50:50 liquid solution containing two components, hexane and nonane You place
    14·1 answer
  • Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: Ca2+, V5+, Br-
    13·1 answer
  • What conclusion can you draw from the experiment about the components of the black ink
    10·1 answer
  • A laser with a wavelength of 225 nm is shown on an isolated gas-phase sodium atom. Calculate the velocity of the ejected electro
    10·1 answer
  • Suppose a 20.0 g gold bar at 35.0°C absorbs 70.0 calories of heat energy. Given that the specific heat of gold is 0.0310 cal/g °
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!