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Sedaia [141]
4 years ago
10

Observation and experimentation have led many scientists to accept a theory about the origin of the universe, called The Big Ban

g Theory. Briefly describe one piece of evidence that supports this scientific theory. (4 points)
Chemistry
2 answers:
34kurt4 years ago
4 0
Red shift is one piece of observational evidence which supports the big bang theory. The fact that galaxies are moving away from us means that the universe was once much smaller, supporting the big bang theory which states that all the matter in the universe once occupied a single point with infinite density.
Anna71 [15]4 years ago
4 0

Answer:

The Big Bang Theory is the dominant cosmological theory about the early development of the universe. Cosmologists use the term "Big Bang" to refer to the idea that the universe was originally very hot and dense at some finite time in the past. Since then it has cooled by expansion to the current diluted state and continues to expand today. The theory is supported by more complete and precise explanations from available scientific evidence and observation. According to the best measurements available in 2010, the initial conditions occurred approximately 13.3 or 13.9 billion years ago.

In general, three empirical evidences that support the cosmological theory of the Big Bang are considered. These are: the expansion of the universe that is expressed in Hubble's law and that can be seen in the redshift of galaxies, the detailed measurements of the cosmic microwave background, and the abundance of light elements. In addition, the correlation function of the large-scale structure of the universe fits with the Big Bang theory.

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In covalent bonds atmosphere share electrons so that?
professor190 [17]
The electons are in the atmosphere
6 0
3 years ago
A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (
djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

Kb = \frac{[ HSO3-][0H-]}{SO3-2}

Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation Kb=\frac{[HSO3-][OH-]}{[SO3-2]}

We noe have:

1.485*10^-^7=\frac{x*x}{(0.056-x)}

x = [OH-] = 9.11*10^-^5

pOH = -log(OH) = -log(9.11*10^-^5)

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0
3 years ago
12. How many grams of glucose (C6H1206) is produced if 17.3 mol of H20 is reacted according to this
iren2701 [21]

Answer:

1. 17.3 MOLES OF WATER WILL PRODUCE 518.4 G OF GLUCOSE.

2. 87.4 G OF HNO3 WILL PRODUCE 7.86 G OF NITROGLYCERIN WHEN REACTED WITH EXCESS GLYCEROL.

Explanation:

1. How many grams of glucose is produced from 17.3 mole of water?

Equation:

6CO2 + 6H20 ------> C6H12O6 + 6O2

From the reaction, 6 moles of carbon dioxide reacts with 6 moles of water to produce 1 mole of glucose

So therefore,

6 moles of water = 1 mole of glucose

Since 17.3 mole of water was used, we can first calculate the number of moles of glucose produced:

Then, we have:

6 moles of water = 1 mole of glucose

17.3 moles of water = ( 17.3 * 1/ 6) moles of glucose

= 2.883 moles of glucose

So we say 17.3 moles of water produces 2.883 moles of glucose

At standard conditions, 1 mole of a substance is its molar mass

Molar mass of water = 18 g/mol

Molar mass of glucose = 180 g/mol

From the reaction:

17.3 moles of water produces 2.883 moles of glucose

17.3 * 18 g of water produces 2.833 * 180 g of glucose

= 518.94 g of glucose.

From 17.3 mole of water, 518.4 g of glucose will be produced.

2.

C3H5(OH)3 + 3HNO3 -------> C3H5(ONO2)3 + 3H20

3 moles of HNO3 reacts to produce 1 mole of nitroglycerin

Molar mass of HNO3 = ( 1 + 14 + 16*3) = 63 g/mol

Molar mass of nitroglycerin = ( 12 *3 + 1 *5 + 16*6 + 14*3) = 179 g/mol

3 moles of HNO3 = 1 mole of nitroglycerin

3 * 63 g of HNO3 = 179 g of nitroglycerin

if 87.4 g of HNO3 were to be reacted, we have:

189 g of HNO3 = 179 g of nitroglycerin

87.4 g of HNO3 = ( 87.4 * 179 / 189) of nitroglycerin

= 7.86 g of nitroglycerin

So therefore, 87.4 g of HNO3 will produce 7.86 g of nitroglycerin when reacted with excess glycerol.

4 0
4 years ago
Suppose that 0.25 mole of gas C was added to
Juli2301 [7.4K]

Answer:

Pi = 0.25[P(TTL)/n(TTL)]

Explanation:

Let Total Pressure = P(TTL) and Total moles =n(TTL) => n(i)/n(TTL) = P(i)/P(TTL)

Given n(i) = 0.25 => Pi = 0.25[P(TTL)/n(TTL)]

6 0
4 years ago
What physical properties is found by dividing the mass of substance by it volume ?
Semenov [28]

Answer:

The physical property is density

5 0
4 years ago
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