The prefix himbo-means heap : true false

lana [24]

c is the answer i odnt know the rest im only a sophmore

3 0

3 years ago

If a balloon initially has a volume of 4.0 liters and a temperature of 10 degrees Celsius, what will the volume of the balloon b

Anika [276]

The answer is D. 8.1L

7 0

2 years ago

1. The heat of fusion for the ice-water phase transition is 335 kJ/kg at 0°C and 1 bar. The density of water is 1000 kg/m3 at th

vodomira [7]

Answer:

Expression for the change of melting temperature with pressure..> T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa), Freezing Point = 0°C

Explanation:

Derivation from state postulate

Using the state postulate, take the specific entropy, , for a homogeneous substance to be a function of specific volume and temperature .

ds = (partial s/partial v)(t) dv + (partial s/partial T)(v) dT

During a phase change, the temperature is constant, so

ds = (partial s/partial v)(T) dv

Using the appropriate Maxwell relation gives

ds = (partial P/partial T)(v) dv

s(β) – s(aplαha) = dP/dT (v(β) – v(α))

dP/dT = s(β) – s(α)/v(β) – v(α) = Δs/Δv

Here Δs and Δv are respectively the change in specific entropy and specific volume from the initial phase α to the final phase β.

For a closed system undergoing an internally reversible process, the first law is

du = δq – δw = Tds - Pdv

Using the definition of specific enthalpy, h and the fact that the temperature and pressure are constant, we have

du + Pdv = dh Tds,

ds = dh/T,

Δs = Δh/T = L/T

After substitution of this result into the derivative of the pressure, one finds

dp/dT = L/TΔv

<u>This last equation is the Clapeyron equation.</u>

a)

(dP/dT) = dH/TdV => dP/dlnT = dH/dV

=> dP/dlnT = dH/dV = [H(liquid) - H(solid)]/[V(liquid) - V(solid)]

= [335,000 J/kg]/[1000⁻¹ - 915⁻¹ m³/kg]

= -3.61x10⁹ J/m³ = -3.61x10⁹ Pa

=> P₂ = P₁ - 3.61x10⁹ ln(T₂/T₁) Pa

or

T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa)

b) if the pressure in Denver is 84.6 kPa:

T₂(freezing) = 273.15exp[-(84,600-100,000)/(3.61x10⁹)]

≅ 273.15 = 0°C T₁(freezing) essentially no change

5 0

3 years ago

What was The US Army trying to slove about lake Lenore in 1947

Leviafan [203]

Answer:

1947 the United States Army had an excess of metallic sodium left over from World War II and determined that the alkaline waters of Lake Lenore would be a good spot to dump and neutralize the acidic element, which reacts with water with intense explosions.

Explanation:

there ya go

5 0

3 years ago

If acetic acid is the only acid that vinegar contains (ka=1.8×10−5), calculate the concentration of acetic acid in the vinegar.

kicyunya [14]

Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.

$CH_{3}COOH \ \textless \ ---\ \textgreater \ H^{+} + CH_{3}COO^{-}$

In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)

pH = 2.88 ==> $[H^{+}]$ = $10^{-2.88}$ = 0.001 $moldm^{-3}$

The change in Concentration Δ $[CH_{3}COOH]$ = 0.001 $moldm^{-3}$

  CH3COOH H+ CH3COOH
Initial   $moldm^{-3}$      $x$ 0 0

Change $moldm^{-3}$ -0.001 +0.001 +0.001

Equilibrium $moldm^{-3}$ $x$ - 0.001 0.001 0.001


Since the $k_{a}$ value is so small, the assumption
$[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium}$ can be made.

$k_{a} = [tex]= 1.8*10^{-5} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} = \frac{0.001^{2}}{x}$

Solve for x to get the required concentration.

note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.

2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind

Hope this helps!

8 0

3 years ago