All categories

4 years ago

Study the diagram above. What cause and effect prediction can be used to test the assumptions in the model?

Chemistry

2 answers:

natima [27]4 years ago

5 0

Answer: alternative D.

Explanation: Condensation consists of decreasing vapor temperature so it becomes liquid once again. Then, when liquid water accumulates, it rains on the system. Precipitation decreases the system temperature due to the increase in air humidity.

VMariaS [17]4 years ago

3 0

Answer: D) condensation will cause the temperature to increase in the system

Explanation: usatestprep approved

You might be interested in

Mention two factors which affect pressure due to liquid contained in a vessel

sveticcg [70]

Answer:

depth of the liquid and nature of liquid affects the pressure due to liquid contained in a vessel

3 0

3 years ago

Read 2 more answers

2. What volume of oxygen is produced at STP when 6.58 x 1024 molecules of water is

astra-53 [7]

Answer:

V = 122.2 L

Explanation:

Given data:

Number of molecules of water = 6.58×10²⁴

Volume of oxygen produced = ?

Solution:

Chemical equation:

2H₂O → 2H₂ + O₂

<em>Number of moles of water:</em>

Number of moles = 6.58×10²⁴/ 6.022 ×10²³

Number of moles = 1.09 ×10¹ mol

Number of moles = 10.9 mol

Now we will compare the moles of water with oxygen:

H₂O : O₂

2 : 1

10.9 : 1/2×10.9 = 5.45 mol

Volume of oxygen:

PV = nRT

V = nRT/P

V = 5.45 mol × 0.0821 L. atm. K⁻¹. mol⁻¹× 273 k / 1 atm

V = 122.2 L. atm. / 1 atm

V = 122.2 L

4 0

4 years ago

PLEASE HELP!! <br><br> Please

dangina [55]

The atomic number 15 is p (phosphorus)
Answer: c

6 0

3 years ago

Consider the reaction 2N2(g) O2(g)2N2O(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for t

ratelena [41]

Answer:

$\Delta G^0 _{rxn} = 207.6\ kJ/mol$

ΔG ≅ 199.91 kJ

Explanation:

Consider the reaction:

$2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}$

temperature = 298.15K

pressure = 22.20 mmHg

From, The standard Thermodynamic Tables; the following data were obtained

$\Delta G_f^0 \ \ \ N_2O_{(g)} = 103 .8 \ kJ/mol$

$\Delta G_f^0 \ \ \ N_2{(g)} =0 \ kJ/mol$

$\Delta G_f^0 \ \ \ O_2{(g)} =0 \ kJ/mol$

$\Delta G^0 _{rxn} = 2 \times \Delta G_f^0 \ N_2O_{(g)} - ( 2 \times \Delta G_f^0 \ N_2{(g)} + \Delta G_f^0 \ O_{2(g)})$

$\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times 0 + 0)$

$\Delta G^0 _{rxn} = 207.6\ kJ/mol$

The equilibrium constant determined from the partial pressure denoted as $K_p$ can be expressed as :

$K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}$

$K_p = \dfrac{1}{ (22.20)}$

$K_p$ = 0.045

$\Delta G = \Delta G^0 _{rxn} + RT \ lnK$

where;

R = gas constant = 8.314 × 10⁻³ kJ

$\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15 \ ln(0.045)$

$\Delta G =207.6 + 2.4788191 \times \ ln(0.045)$

$\Delta G =207.6+ (-7.687048037)$

$\Delta G =$ 199.912952 kJ

ΔG ≅ 199.91 kJ

7 0

4 years ago

O condutor do comboio turístico onde o Francisco viajava a 36 km/h viu um animal da estrada e iniciou uma

Morgarella [4.7K]

I can't understand you what???

Report this if doid not help you

3 0

3 years ago