Formula for Sodium Hydrogen Carbonate NaHCO3? or Na2HCO3?

What is the asteroid's name? a. Bubbles b. Carl c. Bennu d. Rex

castortr0y [4]

Answer:

I would say Bennu

Explanation:

Hope this was correct! Let me know if i need to correct it please =w=

8 0

3 years ago

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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

This last equation is the Clapeyron equation.

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

I need help on dis chem shi if y’all can answer fa me I would appreciate dat

elena55 [62]

I think it’s A I’m not 100% sure but I mean it’s worth a try

3 0

3 years ago

What do these two changes have in common

neonofarm [45]

Both involve chemical bonds breaking

6 0

3 years ago

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Which hydrogen atom would be most easily extracted by a bromine atom?

DerKrebs [107]

The proton which is easily abstracted in 1-Benzyl-3-propylbenzene is the proton which is present on carbon atom in between two phenyl rings, or the central carbon which is shared by two benzene rings.

This easy abstraction of proton is due to its high acidity. Remember those species are always more acidic whose conjugate base is stable. Like the acidity of carboxylic acid is due to stability of the acetate ion.

In our case the stability of conjugate base arises due to stability of negative ion due to resonance. As shown below, the negative charge can delocalize on both rings.

I have shown the resonance of negative ion on both Phenyl rings with Blue and Pink colors.

4 0

3 years ago