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vodomira [7]
1 year ago
6

At which temperature do the molecules of an ideal gas have 3 times the kinetic energy they have at 32of?

Chemistry
1 answer:
algol [13]1 year ago
7 0

Answer:

  • 820 K

Explanation:

As per Boltzman equation, <em>kinetic energy (KE)</em> is in direct relation to the <em>temperature</em>, measured in absolute scale Kelvin.

  • KE α T.

Then, <em>the temperature at which the molecules of an ideal gas have 3 times the kinetic energy they have at any given temperature will be </em><em>3 times</em><em> such temperature.</em>

So, you must just convert the given temperature, 32°F, to kelvin scale.

You can do that in two stages.

  • First, convert 32°F to °C. Since, 32°F is the freezing temperature of water, you may remember that is 0°C. You can also use the conversion formula: T (°C) = [T (°F) - 32] / 1.80

  • Second, convert 0°C to kelvin:

         T (K) = T(°C) + 273.15 K= 273.15 K

Then, <u>3 times</u> gives you: 3 × 273.15 K = 819.45 K

Since, 32°F has two significant figures, you must report your answer with the same number of significan figures. That is 820 K.

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If 200mL of 0.60 M MgCl2 (aq) is added to 400mL of distilled water, what is the concentration of Mg2+(aq) in the resulting solut
lara [203]

Answer:

A.0.20M

Explanation:

c 1 V 1 = c 2 V 2

Initial Volume, V1 = 200 mL

Final Volume, V2 = 200 + 400 = 600 mL

Initial Concentration, c1 = 0.60 M

Final Concentration, c2= ?

Solving for c2;

c2 = c1v1 / v2

c2 = 0.60 * 200 / 600

c2 = 0.20M

3 0
1 year ago
Consider the reaction between nis2 and o2: 2nis2(s)+5o2(g)→2nio(s)+4so2(g) when 11.2 g of nis2 are allowed to react with 5.43 g
stellarik [79]
M(NiS₂) = 11.2 g.
n(NiS₂) = m(NiS₂) ÷ M(NiS₂).
n(NiS₂) = 11.2 g ÷ 122.8 g/mol.
n(NiS₂) = 0.091 mol.
m(O₂) = 5.43 g.
n(O₂) = 5.43 g ÷ 32 g/mol.
n(O₂) = 0.17 mol; limiting reactant.
From chemical reaction: n(NiS₂) : n(O₂) = 2 : 5.
0.091 mol : n(O₂) = 2 : 5.
n(O₂) = 0.2275 mol, not enough.
n(NiO) = 4.89 g .
n(O₂) : n(NiS) = 5 : 2.
n(NiS) = 0.068 mol.
m(NiS) = 0.068 mol · 74.7 g/mol = 5.08 g.
percent yield = 4.89 g / 5.08 g · 100% = 96.2%.


6 0
1 year ago
If 4.168 kJ of heat is added to a calorimeter containing 75.40 g of water, the temperature of the water and the calorimeter incr
Alinara [238K]

Answer:

The value of  the heat capacity of the Calorimeter  C_c = 54.4 \frac{J}{c}

Explanation:

Given data

Heat added Q = 4.168 KJ = 4168 J

Mass of water m_w = 75.40 gm

Temperature change = ΔT = 35.82 - 24.58 = 11.24 ° c

From the given condition

Q = m_w C_w ΔT + C_c ΔT

Put all the values in above equation we get

4168 = 75.70 × 4.18 × 11.24 +  C_c × 11.24

611.37 =  C_c × 11.24

C_c = 54.4 \frac{J}{c}

This is the value of  the heat capacity of the Calorimeter.

7 0
1 year ago
How much energy is transferred when 30.0<br> g of water is cooled from 25.0 °C to 12.7 °C.
enot [183]

Answer:

Heat transferred, Q = 1542.42 J

Explanation:

Given that,

Mass of water, m = 30 grams

Initial temperature, T_i=25^{\circ} C

Final temperature, T_f=12.7^{\circ} C

We need to find the energy transferred. The energy transferred is given by :

Q=mc\Delta T

c is specific heat of water, c = 4.18 J/g °C

So,

Q=30\ g\times 4.18\ J/g-^{\circ} C\times (12.7-25)\ ^{\circ} C\\\\Q=-1542.42\ J

So, 1542.42 J of energy is transferred.

6 0
1 year ago
This same chemistry student has a weight of 155 lbs. What is the student’s weight in
Virty [35]

Answer:

the same chemistry student has a weight of 155lbs what is the student weight in grams? (16 oz= 1 lb, 1 oz= 28.34g)

Explanation:1 lb = 16oz, so multiply your pounds by 16 to get you ounces of the student, then multiply by 28.34 to get grams

155 X 16 X 28.34 = 70283.2

8 0
1 year ago
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