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liubo4ka [24]
3 years ago
11

Water difference between glacier and geyser

Chemistry
2 answers:
Gekata [30.6K]3 years ago
4 0
Glaciers are found in areas where the net accumulation of snow exceeds the melt, therefore they are usually found in areas of high elevation, cold temperatures, and high precipitation.

Geysers are found in areas of shallow crust with high heat flow, such as hot spots.
DENIUS [597]3 years ago
3 0
A geyser a spring characterized by intermittent discharge of water ejected turbulently and accompanied by steam.

glacier: slowly moving mass or river of ice formed by the accumulation and compaction of snow on mountains or near the poles.
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_H2SO4 + __ B(OH)3 = _ B2(SO4)3 + __ H2O
Nana76 [90]

Answer:

3H2SO4 + 2B(OH)3 -> B2(SO4)3 + 6H2O

Explanation:

3 0
3 years ago
A 25.0 ml sample of 0.150 m benzoic acid is titrated with a 0.150 m naoh solution. what is the ph at the equivalence point? the
Ad libitum [116K]

Solution:

At the equivalence point, moles NaOH = moles benzoic acid  

HA + NaOH ==> NaA + H2O where HA is benzoic acid  

At the equivalence point, all the benzoic acid ==> sodium benzoate  

A^- + H2O ==> HA + OH- (again, A^- is the benzoate anion and HA is the weak acid benzoic acid)  

Kb for benzoate = 1x10^-14/4.5x10^-4 = 2.22x10^-11  

Kb = 2.22x10^-11 = [HA][OH-][A^-] = (x)(x)/0.150  

x^2 = 3.33x10^-12  

x = 1.8x10^-6 = [OH-]  

pOH = -log [OH-] = 5.74  

pH = 14 - pOH = 8.26  


5 0
3 years ago
Read 2 more answers
Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi
Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g)     ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

[]=\frac{P}{R.T}

<em>Calculate ΔG at 25°C given the following sets of partial pressures.</em>

<em>Part A  130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.</em>

[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M

[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

<em>Part B  5.0atm SO₂, 3.0atm O₂, 30atm SO₃  Express your answer using four significant figures.</em>

<em />

[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M

[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M

[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

<em>Part C Each reactant and product at a partial pressure of 1.0 atm.  Express your answer using four significant figures.</em>

<em />

[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0
2 years ago
why is it better to use several trials of a titration rather than one trial to determine the concentration of a solution of unkn
svlad2 [7]
Because you are never 100% precise during the work, so it's best and most accurate answer is always the average of more trials. basically the more you do, the more accurate the answer shall be 
6 0
2 years ago
ANSWER FAST PLZ 25 POINTS!!!!!!!!!!!!!!!!
Rainbow [258]

Answer:

Point Z

Explanation:

The river flows this direction pushing the water pollution to Point Z

3 0
3 years ago
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