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3 years ago

Which description is of the man in the graveyard?

Chemistry

2 answers:

Mashutka [201]3 years ago

6 0

Answer: hungry

Explanation:

vekshin13 years ago

5 0

B is the answer to ur question

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Describe the properties of oobleck when it behaves like a solid and a liquid.

PSYCHO15rus [73]

Answer:

hope it's help you ok have a good day

7 0

3 years ago

Give me lil reasoning so I know your not lying for points

jeka94

Answer:

0.01 psi

Explanation:

If you look at the data points plotted on the graph, the slope of the line touches 0.1 for the y-axis when it is at 20 for the x-axis.

3 0

3 years ago

The temperature is -14°C the air pressure in an automobile tire is 149K PA if the volume does not change what is the pressure af

Jet001 [13]

Answer: The pressure after the tire is heated to 17.3°C is 167 kPa

Explanation:

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=149kPa\\T_1=-14^0C=(273-14)=259K\\P_2=?=27.5psi\\T_2=17.3^0C=(273+17.3)=290.3K$

Putting values in above equation, we get:

$\frac{149}{259}=\frac{P_2}{290.3}\\\\P_2=167kPa$

Hence, the pressure after the tire is heated to 17.3°C is 167 kPa

4 0

3 years ago

A mixture initially contains AA, BB, and CC in the following concentrations: [A][A]A_1 = 0.550 MM , [B][B]B_1 = 1.40 MM , and [C

Alex787 [66]

Answer:

The value of the equilibrium constant KC is 1.244

Explanation:

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.550 M, [B] = 1.40 M, and [C] = 0.600 M. The following reaction occurs and equilibrium is established: A+2B<->C

At equilibrium, [A] = 0.430 M and [C] = 0.720 M. Calculate the value of the equilibrium constant, Kc

Step 1: The balanced equation

A+2B<->C

Step 2: The initial concentrations

[A] = 0.550 M

[B]= 1.40 M

[C] = 0.600 M

Step 3: The concentraions at equilibrium

[A] = 0.550 -X = 0.430 M

[B]= 1.40 -2X M

[C] = 0.600 + X = 0.720 M

X = 0.120 M

[A] = 0.550 - 0.120 = 0.430 M

[B]= 1.40 -2*0.120 = 1.16 M

[C] = 0.600 + 0.120 = 0.720 M

Step 4: Calculate Kc

Kc = [C] / [A][B]²

Kc = 0.720 / (0.430*1.16²)

Kc = 1.244

The value of the equilibrium constant KC is 1.244

5 0

3 years ago

According to reference table adv-10, which reaction will take place spontaneously?

olga_2 [115]

Missing table!! write the elements with the first letter of the symbol with Upper Caps letters!!!

http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm

Ni2+ +Pb(s) → Ni(s) + Pb2+
The potential of the oxidation of Pb(s) --> Pb2+(aq) is 0.126 V
The potential of the reduction go Ni2+(aq) --> Ni(s) is -0.25 V

Add the two together and the potential for the reaction is -0.124 V (NO SPONTANEOUS THE SIGN IS NEGATIVE)

au3+ + al(s) → au(s) + al3+Au3+(aq) -> Au(s) +1.5 VAl -> Al3+ +1.66VV= 3.16 (SPONTANEOUS THE SIGN OF THE PONTENTIAL IS POSITIVE)Sr2+ + Sn(s) → Sr(s) + Sn2+

Sr2+(aq) + 2 e– Sr(s) V= -2.89V
Sn -> Sn2+ V= 0.14 V
V= -2.75 V (no spontaneous)

Fe2+ + Cu(s) → Fe(s) + Cu2+
Fe2+(aq) + 2 e– Fe(s) V= -0.44 V
Cu -> C2+ V = - 0.337V

V= - 0.777V (no spontaneous)

5 0

3 years ago