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Kisachek [45]
3 years ago
12

How much H2O is produced when 20000 g of C2H2 burns completely? Answer in units of g.

Chemistry
1 answer:
IgorC [24]3 years ago
3 0

Answer:

Explanation:

H  = 1

C = 12

O = 16

Acetylene, HC≡CH = 2+24 = 26

H2O = 2 + 16 = 18

In XS oxygen, one HC≡CH yields one H2O

26 g HC≡CH  ==> 18 g H2O

2000 g HC≡CH ==> 2000*18/26 g H2O =  1384.6154 g H2O

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yKpoI14uk [10]

Answer:

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Hope This Helps!

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3 years ago
If a person did overshoot the endpoint, would this cause the experimental value for the % of acid in vinegar to be too high or t
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3 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
An unknown noble gas has a density of 5.84 g dm-3 at STP. Calculate its molar mass, and so identify the gas.
bagirrra123 [75]

The noble gas is Xenon and its molar mass is 131 g/mol.

<h3>What is the molar mass of the noble gas?</h3>

The molar mass of the noble gas is determined as follows;

Let molar mass of unknown gas be M, and mass of gas be m

Density of the noble gas, ρ = 5.8 g/dm³

density = m/V

At STP;

  • temperature, T = 273.15 K
  • pressure, P = 1 atm
  • molar gas constant, R = 0.0821 L.atmK⁻¹mol⁻¹

From ideal gas equation:

PV = nRT

where n = m/M

PV = mRT/M

M = mRT/PV

M =  0.0821 * 273.15 * 5.84/1

Molar mass of the noble gas = 131 g/mol

The noble gas is Xenon which has molar mass approximately equal to 131 g/mol.

Learn more about molar mass at: brainly.com/question/837939

#SPJ1

6 0
1 year ago
In the presence of a base, blue litmus paper will .<br>O turn purple<br>stay blue<br>to turn red​
Inessa05 [86]

Answer:

In the presence of a base, blue litmus paper will turn red........

7 0
3 years ago
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