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krek1111 [17]
3 years ago
15

What does not release carbon dioxide into the atmosphere

Chemistry
2 answers:
Galina-37 [17]3 years ago
6 0
<span>There are both natural and human sources of carbon dioxide emissions. Natural sources include decomposition, ocean release and respiration. Human sources come from activities like cement production, deforestation as well as the burning of fossil fuels like coal, oil and natural gas. </span>
frez [133]3 years ago
6 0

yo mama doesn't you big headed person


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Calculează cantitatea de zahăr (C12 H22 O11) în ceaiul de dimineața dacă o linguriță conține 5g de zahăr?
slavikrds [6]

Answer:

Explanation:

Translated:

Calculate the amount of sugar (C12 H22 O11) in the morning tea if a teaspoon contains 5g of sugar?

Given parameters

Mass of sugar = 5g

Unknown:

Amount of the sugar =?

Solution:

A mole is the amount of substance that contains avogadro's number of particles i.e 6.02 x 10²³

Therefore, to find the number of moles contained in the teaspoon of sugar we use the expression below:

    Number of moles = \frac{mass}{molar mass}

Molar mass of C₁₂H₂₂O₁₁ = (12x12) + (1x22) + (16x11) = 342g/mol

    Number of moles = \frac{5}{342} = 0.015mol of sugar

or 0.015 x 6.02 x 10²³, 8.8x10²¹atoms of sugar

8 0
3 years ago
Calculate the standard entropy change for the following reactions at 25°C.
Art [367]

Answer:

(a) ΔSº = 216.10 J/K

(b) ΔSº = - 56.4 J/K

(c) ΔSº = 273.8 J/K

Explanation:

We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.

First we need to find in an appropiate reference table the standard  molar entropies entropies, and then do the calculations.

(a)        C2H5OH(l)          +        3 O2(g)         ⇒        2 CO2(g)     +    3 H2O(g)

Sº            159.9                          205.2                         213.8                  188.8

(J/Kmol)

ΔSº = [ 2(213.8) + 3(188.8) ]   - [ 159.9  + 3(205.) ]  J/K

ΔSº = 216.10 J/K

(b)         CS2(l)               +         3 O2(g)               ⇒      CO2(g)      +      2 SO2(g)

Sº          151.0                              205.2                         213.8                 248.2

(J/Kmol)

ΔSº  = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K

(c)        2 C6H6(l)           +        15 O2(g)                     12 CO2(g)     +     6 H2O(g)

Sº           173.3                           205.2                           213.8                    188.8

(J/Kmol)  

ΔSº  = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K

Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4  total mol gas reactants to 3, so the entropy change will be negative.

Note we need to multiply the entropies of each substance by  its coefficient in the balanced chemical equation.

5 0
3 years ago
A Bronsted- Lowry base
kari74 [83]
A bronsted lowry base will react to accept protons
3 0
3 years ago
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Number 2 I don't get what the question is asking
Anettt [7]
Can you post the question, on here? I cant open the document. 
3 0
3 years ago
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Consider the two triangles shown below.
attashe74 [19]

Triangles may be regarded as similar or congruent on the following basis;

1) Side angle side (SAS)

2) Side side side (SSS)

3) Angle Angle side (AAS)

<h3>What is a triangle?</h3>

A triangle is a three sided figure. The figures are not shown here. However, two triangle may be regarded as similar or congruent by the following conditions;

1) Side angle side (SAS)

2) Side side side (SSS)

3) Angle Angle side (AAS)

Since the triagles are not shown here, the similarity of the triangles can not be established.

Learn more about triangles: brainly.com/question/25813512

8 0
2 years ago
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