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igomit [66]
3 years ago
15

Why is it important for a scientist studying greenhouse effect to know about mater,forces and energy

Chemistry
1 answer:
vova2212 [387]3 years ago
8 0
<span>In the 19th century, scientists realized that gases in the atmosphere cause a "greenhouse effect" which affects the planet's temperature. These scientists were interested chiefly in the possibility that a lower level of carbon dioxide gas might explain the ice ages of the distant past. At the turn of the century, Svante Arrhenius calculated that emissions from human industry might someday bring a global warming. Other scientists dismissed his idea as faulty. In 1938, G.S. Callendar argued that the level of carbon dioxide was climbing and raising global temperature, but most scientists found his arguments implausible. It was almost by chance that a few researchers in the 1950s discovered that global warming truly was possible. In the early 1960s, C.D. Keeling measured the level of carbon dioxide in the atmosphere: it was rising fast. Researchers began to take an interest, struggling to understand how the level of carbon dioxide had changed in the past, and how the level was influenced by chemical and biological forces. They found that the gas plays a crucial role in climate change, so that the rising level could gravely affect our future. (This essay covers only developments relating directly to carbon dioxide, with a separate essay for Other Greenhouse Gases. Theories are discussed in the essay on Simple Models of Climate.)</span>
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What is the result when earth layers experience extension stress? Question 2 options:
Vika [28.1K]

Answer:

stretching and thinning of the Earth's crust occurs.

Explanation:

The result of the extension stress on Earth layers is the stretching and thinning of the Earth's crust occurs.

When there is a stress on the Earth's layers, the Earth's crust experiences the phenomena of stretching and thinning.

4 0
2 years ago
density of a piece of ebony wood is 1.174 g/ml. what is the volume in quarts of a 2.1549 lb piece of this ebony wood?
lapo4ka [179]

0.88qt

Explanation:

Given parameters:

Density of ebony wood = 1.174g/ml

Mass of wood = 2.1549lb

Unknown:

Volume of the wood in quarts = ?

Solution:

Density is defined as the mass per unit volume of a substance.

Mass is the quantity of matter the wood contains.

Volume is the space it occupies.

To solve this problem, convert the mass form pounds into grams and plug into the density equation to find volume.

Then convert the volume from mL to quarts.

 Density = \frac{mass}{volume}

 Volume = \frac{mass}{density}

    1 pound = 453.592g

   2.1549 lb =  2.1549 x 453.592 = 977.5g

Volume = \frac{977.5}{1.174} = 832.6mL

     1 quart = 946.353mL

        Therefore 832.6mL x \frac{1qt}{946.353} = 0.88qt

learn more:

Density brainly.com/question/8441651

#learnwithBrainly

7 0
3 years ago
I want to know the steps.
Artyom0805 [142]

The answer for the following problem is described below.

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

Explanation:

Given:

enthalpy of combustion of glucose(ΔH_{f} of C_{6}H_{12} O_{6}) =-1275.0

enthalpy of combustion of oxygen(ΔH_{f} of O_{2}) = zero

enthalpy of combustion of carbon dioxide(ΔH_{f} of CO_{2}) = -393.5

enthalpy of combustion of water(ΔH_{f} of H_{2} O) = -285.8

To solve :

standard enthalpy of combustion

We know;

ΔH_{f}  = ∈ΔH_{f} (products) - ∈ΔH_{f} (reactants)

C_{6}H_{12} O_{6} (s) +6 O_{2}(g) → 6 CO_{2} (g)+ 6 H_{2} O(l)

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

ΔH_{f} = 6 (-393.5) + 6(-285.8)  - 0 + 1275

ΔH_{f} = -2361 - 1714 - 0 + 1275

ΔH_{f} =-2800 kJ

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

7 0
3 years ago
Write a hypothesis using IF-THEN - BECAUSE of statement. (PLEASE HELP)
Mekhanik [1.2K]

Water is an essential part of life and its availability is important for all living creatures. On the other side, the world is suffering from a major problem of drinking water. There are several gases, microorganisms and other toxins (chemicals and heavy metals) added into water during rain, flowing water, etc. which is responsible for water pollution. This review article describes various applications of nanomaterial in removing different types of impurities from polluted water. There are various kinds of nanomaterials, which carried huge potential to treat polluted water (containing metal toxin substance, different organic and inorganic impurities) very effectively due to their unique properties like greater surface area, able to work at low concentration, etc. The nanostructured catalytic membranes, nanosorbents and nanophotocatalyst based approaches to remove pollutants from wastewater are eco-friendly and efficient, but they require more energy, more investment in order to purify the wastewater. There are many challenges and issues of wastewater treatment. Some precautions are also required to keep away from ecological and health issues. New modern equipment for wastewater treatment should be flexible, low cost and efficient for the commercialization purpose.

3 0
3 years ago
How many grams of calcium phosphate (Ca3(PO4)2) re theoretically produced if we start with 3.40 moles of Ca(NO3)2 and 2.40moles
sattari [20]

1) Balance the chemical equation.

3Ca(NO_3)_2+2Li_3PO_4\rightarrow6LiNO_3+Ca_3(PO_4)_2

2) List the known and unknown quantities.

Reactant 1: Ca(NO3)2.

Amount of substance: 3.40 mol.

Reactant 2: Li3PO4.

Amount of substance: 2.40 mol.

Product: Ca3(PO4)2

Mass: unknown.

3) Which is the limiting reactant?

<em>3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?</em>

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

mol\text{ }Li_3PO_4=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{2\text{ }mol\text{ }Li_3PO_4}{3\text{ }mol\text{ }Ca(NO_3)_2}=2.2667\text{ }mol\text{ }Li_3PO_4

<em>We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. </em>This is the excess reactant.

<em>3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?</em>

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

mol\text{ }Ca(NO_3)_2=2.40\text{ }mol\text{ }Li_3PO_4*\frac{3\text{ }mol\text{ }Ca(NO_3)_2}{2\text{ }mol\text{ }Li_3PO_4}=3.60\text{ }mol\text{ }Ca(NO_3)_2

<em>We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. </em>This is the limiting reactant.

4) Moles of Ca3(PO4)2 produced from the limiting reactant.

We have 3.40 mol Ca(NO3)2 of the limiting reactant.

The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.

mol\text{ }Ca_3(PO_4)_2=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{1\text{ }mol\text{ }Ca_3(PO_4)_2}{3\text{ }mol\text{ }Ca(NO_3)_2}=1.1313\text{ }mol\text{ }Ca_3(PO_4)_2

5) Mass of Ca3(PO4)2 produced.

The molar mass of Ca3(PO4)2 is 310.1767 g/mol.

g\text{ }Ca_3(PO_4)_2=1.1333\text{ }mol\text{ }Ca_3(PO_4)_2*\frac{310.1767\text{ }g\text{ }Ca_3(PO_4)_2}{1\text{ }mol\text{ }Ca_3(PO_4)_2}g\text{ }Ca_3(PO_4)_2=351.526\text{ }g\text{ }Ca_3(PO_4)_2

<em>The mass of Ca3(PO4)2 produced is</em> 351 g Ca3(PO4)2.

Option D.

.

8 0
1 year ago
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