What is the starting molecule for glycolysis?

The structure of haloacetic acids, XCH,COOH (where X is either F.CI, Br, or I), is shown above. The dissociation constants and m

lana [24]

Answer:

A student titrates 20.0mL of 1.0M NaOH with 2.0M formic acid, HCO2H (Ka=1.8x10-4). Formic ... How much formic acid is necessary to reach the equivalence point? a. ... At the equivalence point, is the solution acidic, basic or neutral? ... By adding a small amount of water to the beaker but not dissolving all of the solid. 11.

Explanation:

8 0

2 years ago

A rigid, 28-L steam cooker is arranged with a pressure relief valve set to release vapor and maintain the pressure once the pres

MaRussiya [10]

Answer:

$\Delta S_{source}>-1.204\frac{kJ}{K}$

Explanation:

Hello!

In this case, given the initial conditions, we first use the 10-% quality to compute the initial entropy:

$s_1=s_{f,175kPa}+q*s_{fg,175kPa}\\\\s_1=1.4850\frac{kJ}{kg*K} +0.1*5.6865\frac{kJ}{kg*K}=2.0537\frac{kJ}{kg*K}$

Now the entropy at the final state given the new 40-% quality:

$s_2=s_{f,150kPa}+q*s_{fg,150kPa}\\\\s_2=1.4337\frac{kJ}{kg*K} +0.4*5.7894\frac{kJ}{kg*K}=3.7495\frac{kJ}{kg*K}$

Next step is to compute the mass of steam given the specific volume of steam at 175 kPa and the 10% quality:

$m_1=\frac{0.028m^3}{(0.001057+0.1*1.002643)\frac{m^3}{kg} } =0.274kg\\\\m_2=\frac{0.028m^3}{(0.001053+0.4*1.158347)\frac{m^3}{kg} } =0.0603kg$

Then, we can write the entropy balance:

$\Delta S_{source}+\frac{Q}{T_1} -\frac{Q}{T_2} +s_2m_2-s_1m_1-s_{fg}(m_2-m_1)>0$

Whereas sfg stands for the entropy of the leaving steam to hold the pressure at 150 kPa and must be greater than 0; thus we plug in:

Which is such minimum entropy change of the heat-supplying source.

Best regards!

3 0

2 years ago

Some instant cold packs contain ammonium nitrate and a separate pouch of water. When the pack is activated by squeezing to break

lubasha [3.4K]

Answer:

$T_f=-7.81^0C$

Explanation:

Hello,

a) In this case, since the heat associated with the dissolution of ammonium nitrate is positive, such reaction is endothermic as it absorbs heat.

b) Now, for computing the temperature once the dissolution is done, we apply (considering that it is a cooling process):

$T_f=T_0-\frac{\Delta H}{mCp}$

Nonetheless, we should first compute the moles of the mixture as:

$n_{mix}=135.0gH_2O*\frac{1molH_2O}{18gH_2O}+50.0gNH_4NO_3*\frac{1molNH_4NO_3}{80gNH_4NO_3}=8.125mol$

Thus, the total absorbed heat is:

$\Delta H=25.4kJ/mol*8.125mol=206.375kJ$

Now, the temperature is:

$T_f=25.0^0C-\frac{25.4kJ}{(135.0+50.0)g*4.184x10^{-3}kJ/g^0C} \\\\T_f=-7.81^0C$

Best regards.

3 0

2 years ago

When 50 ml (50 g) of 1.00 m hcl at 22oc is added to 50 ml (50 g) of 1.00 m naoh at 22oc in a coffee cup calorimeter, the tempera

vitfil [10]

Answer:

$\boxed{\text{2700 J}}$

Explanation:

HCl + NaOH ⟶ NaCl + H₂O

There are two energy flows in this reaction.

Heat of reaction + heat to warm water = 0

q₁ + q₂ = 0

q₁ + mCΔT = 0

Data:

m(HCl) = 50 g

m(NaOH) = 50 g

T₁ = 22 °C

T₂ = 28.87 °C

C = 4.18 J·°C⁻¹g⁻¹

Calculations:

m = 50 + 50 = 100 g

ΔT = 28.87 – 22 = 6.9 °C

q₂ = 100 × 4.18 × 6.9 = 2900 J

q₁ + 2900 = 0

q₁ = -2900 J

The negative sign tells us that the reaction produced heat.

The reaction produced $\boxed{\textbf{2900 J}}$ .

7 0

2 years ago

Consider the following balanced thermochemical equation for a reaction sometimes used for H2S production:

fgiga [73]

Answer:

d. Heat is released from the reaction

Explanation:

A negative enthalpy change indicates that it is an exothermic reaction. Exothermic reactions release heat.

5 0

2 years ago