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2 years ago

An ice cube is dropped into a glass of milk at 20°C. Which statement explains the transfer of kinetic energy (KE)?

Chemistry

1 answer:

hoa [83]2 years ago

3 0

Answer:

Because milk has higher KE than ice, KE is transferred from the milk to the molecules of ice.

Explanation:

The best statement that expresses the transfer of kinetic energy(K.E) is that kinetic energy is transferred from the milk to the ice.

Kinetic energy is form of energy due to motion of the particles of a medium. In this regard, we are dealing with heat energy.

Heat energy is dissipated from a body at higher temperature to one at a lower temperature.
Ice is at a lower temperature which is 0°C
Heat will be transferred in form of thermal energy from the body at higher temperature to one with a lower temperature.
This is from the milk to the molecules of ice.

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For the reaction 3h2(g) + n2(g) 2nh3(g), kc = 9.0 at 350°c. calculate g° at 350°c.

miss Akunina [59]

When ΔG° is the change in Gibbs free energy

So according to ΔG° formula:

ΔG° = - R*T*(㏑K)

here when K = [NH3]^2/[N2][H2]^3 = Kc

and Kc = 9

and when T is the temperature in Kelvin = 350 + 273 = 623 K

and R is the universal gas constant = 8.314 1/mol.K

So by substitution in ΔG° formula:

∴ ΔG° = - 8.314 1/ mol.K * 623 K *㏑(9)

= - 4536

4 0

2 years ago

6. Un volumen de 1.0 mL de agua de mar contiene casi 4 x 10-12 g de Au. El volumen total de agua en los océanos es de 1.5 x 1021

Aleks [24]

Answer:

The total amount of Au is $ $2.0\times10^{24}$

Explanation:

Given that,

Mass of 1.0 ml of Au $m=4\times10^{-2}\ g$

Total volume of water in oceans $V=1.5\times10^{21}\ L$

We need to calculate the volume in ml

Using given volume

$V=1.5\times10^{21}\times1000\ mL$

$V=1.5\times10^{24}\ mL$

We need to calculate the total mass of Au

Using given data

$1\ ml\ volume = 4\times10^{-2}\ g$

$1.5\times10^{24}\ ml=4\times10^{-2}\times1.5\times10^{24}$

So, The total mass of Au is $6\times10^{22}\ g$

The mass will be in ounce,

$Mass=0.035274\times6\times10^{22}$

$Mass=2.12\times10^{21}\ ounce$

The total amount of the Au Will be

$Total\ amount=2.12\times10^{21}\times948$

$Total\ amount=2.0\times10^{24}$

Hence, The total amount of Au is $ $2.0\times10^{24}$

3 0

2 years ago

How many sigma and pi bonds in propionic bond

Ierofanga [76]

Propanoic acid formula is ch ch 2 so it has 8 bonds

7 0

2 years ago

Acetone has a boiling point of 56.5 celcius. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57

vodomira [7]

Answer:

0.515 g

Explanation:

<em>Acetone (C₃H₆O) has a boiling point of 56.5 °C. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57 °C and 730 mmHg?</em>

<em />

Step 1: Given data

Temperature (T): 57°C

Pressure (P): 730 mmHg

Volume (V): 250 mL

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15 = 57°C + 273.15 = 330 K

Step 3: Convert "P" to atm

We will use the conversion factor 1 atm = 760 mmHg.

730 mmHg × (1 atm/760 mmHg) = 0.961 atm

Step 4: Convert "V" to L

We will use the conversion factor 1 L = 1,000 mL.

250 mL × (1 L/1,000 mL) = 0.250 L

Step 5: Calculate the moles (n) of acetone

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 0.961 atm × 0.250 L/(0.0821 atm.L/mol.K) × 330 K

n = 8.87 × 10⁻³ mol

Step 6: Calculate the mass corresponding to 8.87 × 10⁻³ moles of acetone

The molar mass of acetone is 58.08 g/mol.

8.87 × 10⁻³ mol × 58.08 g/mol = 0.515 g

8 0

2 years ago

Carbonic acid, H2CO3, has two acidic hydrogens. A solution containing an unknown concentration of carbonic acid is titrated with

stepan [7]

Answer:

1) Net ionic equation :

$2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)$

2) 0.765 M is the molarity of the carbonic acid solution.

Explanation:

1) In aqueous carbonic acid , carbonate ions and hydrogen ion is present.:

$H_2CO_3(aq)\rightarrow 2H^+(aq)+CO_3^{2-}(aq)$ ..[1]

In aqueous potassium hydroxide , potassium ions and hydroxide ion is present.:

$KOH(aq)\rightarrow K^+(aq)+OH^{-}(aq)$ ..[2]

In aqueous potassium carbonate , potassium ions and carbonate ion is present.:

$K_2CO_3(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)$ ..[3]

$H_2CO_3(aq)+2KOH(aq)\rightarrow K_2CO_3(aq)+2H_2O(l)$

From one:[1] ,[2] and [3]:

$2H^+(aq)+CO_3^{2-}(aq)+2K^+(aq)+2OH^{-}(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)+H_2O(l)$

Cancelling common ions on both sides to get net ionic equation :

$2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)$

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2CO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=2\\M_1=?\\V_1=50.0 mL\\n_2=1\\M_2=3.840 M\\V_2=20.0 mL$

Putting values in above equation, we get:

$M_1=\frac{1\times 3.840 M\times 20.0 mL}{2\times 50.2 mL}=0.765 M$

0.765 M is the molarity of the carbonic acid solution.

6 0

2 years ago