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astra-53 [7]
2 years ago
11

Provide an example of at least one interaction for each pair of spheres. 1. lithosphere and atmosphere: 2. lithosphere and hydro

sphere: 3. lithosphere and biosphere: 4. hydrosphere and atmosphere: 5. hydrosphere and biosphere: 6. atmosphere and biosphere:
Chemistry
1 answer:
Sidana [21]2 years ago
3 0

Answer:

  • 1. lithosphere and atmosphere:  Rocks and rain.
  • 2. lithosphere and hydrosphere:  Oceans and seas.
  • 3. lithosphere and biosphere:  Wildlife and natural vegetation.
  • 4. hydrosphere and atmosphere: insolation and sea level rise.
  • 5. hydrosphere and biosphere: Mangrove forests and swamps.
  • 6. atmosphere and biosphere: GHG and energy flows.

Explanation:

  • The example pf the land and the air is that of soil, rocks, and the rainwater that falls from the clouds. The example of the land and water is that of the seas and the oceans and other water bodies. The example of the land and life sphere is the flora and fauna.  
  • The example of the water and air is that of insulation, carbon cycles, and a rise in the sea levels due to ice melt. Water and biosphere include the mangrove forest s and swamps. The air and biosphere include the examples of greenhouse effect and energy flows in an ecosystem.
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slava [35]

Answer:

(a) A strong acid

Explanation:

We have given the pH of the solution is 2.46

pH=2.46  

So the concentration of H^+=10^{-pH}=10^{-2.46}=0.00346

solution having H+ concentration more than H^+=10^{-7} is acidic

Since in the given solution, H+ concentration is 0.00346 M which is more than 10^{-7}[/tex] so this is an acidic solution

Note-The concentration of H^+ decide the behavior of the solution that is, it is acidic or basic

7 0
2 years ago
A 6.0M solution HCl is diluted to 1.0M How many milliliters of the 6.0M solution would be used to prepare 100.o mL of the dilute
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An unknown compound melting at 131 - 133 C. It is thought to be one of the following compounds: trans-cinnamic acid (133-134); b
Vesnalui [34]

Answer:

benzamide

Explanation:

Compound            melting Point ,ºC          Melting Pont Mixture, ºC

       X                          131 - 133

trans-cinnamic            133 - 134                      110 - 120

acid

benzamide                 128 - 130                       130-132

malic acid                   131   -133                        114 -124

Benzoin                      135 - 137                        108 - 116

The compound X is benzamide since the melting point range is the one closest to this compound (  130-132 ºC)

The reason there is not an exact match is not due due to the presence of impurities. The presence of impurities always lower the melting point ( it is a coligative property such as the melting point depresion of salt and water )

The reason for the deviation must be  be some other factors such as preparation of the sample in the capillary, errors in reading the thermometer, rate of heating, etc.

5 0
2 years ago
HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.
Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

concentration of HA, C_A = 6.0mol/dm³

volume of HA, V_A  = 25.0cm³, = 0.025dm³

Concentration of HB, C_B = 3.0mol/dm³

volume of HB, V_B = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

C_iVi = C_fV_f

where;

C_i is initial concentration

V_i is initial volume

C_f is final concentration of the solution

V_f is final volume of the solution

C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3

Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

7 0
2 years ago
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AleksAgata [21]

Answer:

The true statement  is option A.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas =  1 atm

V = Volume of gas = ?

n = number of moles of gas = 1 mol

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 273.15 K

V=\frac{nRT}{P}=\frac{1 mol\times 0.0821 atm L/mol K\times 273.15 K}{1 atm}

V = 22.42 L

This means that 1 mole of an ideal gas at STP occupies 22.42 liters of volume.

So, 1 mole of helium gas and 1 mole of oxygen gas will have same value of volume in their respective balloons at STP.

7 0
2 years ago
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