Are spider multicellular

Express in scientific notation. Make sure your answer has the same number of significant figures as the starting value. (1.8 x 1

slamgirl [31]

Answer : The answer is, $0.2\times 10^{-4}$

Explanation :

Scientific notation : It is defined as the way or representation of expressing the number that are too big or too small that is written in the decimal form. That means always written in the power of 10 form.

For example : 8000 is written as, $8.0\times 10^3$ in this notation, the significant figure is, 1 but $0.8\times 10^4$ in this notation, the significant figures are, 2

The given problem are :

$\frac{1.8\times 10^{-2}}{9\times 10^2}=0.2\times 10^{-4}$

According to the divide and multiplication rules of scientific figures, the final answer is only as precise as the value with the least number of significant figures.

Hence, the answer is, $0.2\times 10^{-4}$

6 0

3 years ago

Read 2 more answers

Convert 32 Fahrenheit to Celsius

valina [46]

Answer:

°F - 32/1.8000

Explanation:

8 0

3 years ago

The equilibrium constant in terms of pressure. Kp for the following process is 0.179 at 50 °C. It increases to 0.669 at 86 °C. C

Vikentia [17]

Answer:

a) ΔHvap=35.3395 kJ/mol

b) Tb=98.62 °C

Explanation:

Given the reaction:

C₇H₁₆ (l) ⇔ C₇H₁₆ (g)

Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.

When

T₁=50°C=323.15K ⇒P₁=0.179

T₂=86°C=359.15K ⇒P₂=0.669

The Clasius-Clapeyron equation is:

$ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})$

$ln(\frac{0.669}{0.179}) =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{359.15K}-\frac{1}{323.15K})$

$1.3184 =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (-3.10186*10^{-4}K{^-1})$

ΔHvap=35339.5 J/mol=35.3395 KJ/mol

Normal boiling point ⇒ P=1 atm

Hence, we find the normal boiling point where:

T₁=323.15K

P₁=0.179 atm

P₂=1 atm

$ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})$

$ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})$

$1.7203=-4250.34 (\frac{1}{T_2}-\frac{1}{323.15K})$

T₂=371.77 K= 98.62 °C

5 0

3 years ago

1. Show that heat flows spontaneously from high temperature to low temperature in any isolated system (hint: use entropy change

Inga [223]

Answer:

1 ) Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

= Q/tc - Q/th

2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

Explanation:

1) To show that heat flows spontaneously from high temperature to low temperature

example :

Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )

Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment

Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )

Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

= Q/tc - Q/th

2) Entropy change for Decomposition of mercuric oxide

2HgO (s) → 2Hg(l) + O₂ (g)

Δs = positive

there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C

hence ΔSdecomposition = S⁻ Hg - S⁻ HgO =

Δh of reaction = 181.6 KJ

Temp = 500 + 273 = 773 k

hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

8 0

3 years ago

Which of the following is not a postulate of the kinetic molecular theory?

tester [92]

Explanation:

Which of the following is not a postulate of the kinetic molecular theory?

Answer is

option D. the collisions between gas molecules are elastic

Hope it will help you

6 0

3 years ago