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Scrat [10]
3 years ago
14

Fill in the coefficients that will balance the following reaction a0h2so4 + a1koh → a2k2so4 + a3hoh

Chemistry
2 answers:
Anuta_ua [19.1K]3 years ago
7 0
A0 = 1
a1 = 2
a2 = 1
a3 = 2

This can be solved by guessing and checking and making sure that all atoms on the left side are accounted for on the right side. Both sides must have the same amount of atoms or the balancing is not correct.
iris [78.8K]3 years ago
6 0

a0h2so4 + a1koh → a2k2so4 + a3hoh

The balanced chemical reaction is:

H2SO4 + 2KOH → K2SO4 + 2H2O


hence,

a0 = 1

a1 = 2

a2 = 1

a3 = 2

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Each liter of air has a mass of 1.80 grams. How many liters of air are contained in
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25kg

Explanation:

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Define "Mutarotation"
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Answer:

Mutarotation refers to the change in the optical rotation or optical activity of a solution due to the change in the equilibrium of the two anomers. It depends upon the optical activity and ratio of the anomeric forms in the solution.

To measure the optical rotation of a given solution, a polarimeter can be used and thus the ratio of the anomeric forms can be calculated.

3 0
2 years ago
calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w
kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is -22Jmole^-^1k^-^1

(b) The normal boiling point of water (J·K−1·mol−1) is -109Jmole^-^1K^-^1

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is  109J/mole

Explanation:

Lets calculate

(a) - General equation -

      (\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p = -5_m(\beta )+5_m(\alpha ) =  -\frac{\Delta H}{T}

 \alpha ,\beta → phases

ΔH → enthalpy of transition

T → temperature transition

 (\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p == -\frac{\Delta_fH}{T_f}

            = \frac{-6.008kJ/mole}{273.15K} ( \Delta_fH is the enthalpy of fusion of water)

           = -22Jmole^-^1k^-^1

(b) (\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}

                                  = \frac{40.656kJ/mole}{373.15K} (\Delta_v_a_p_o_u_rH is the enthalpy of vaporization)

                               = -109Jmole^-^1K^-^1

(c) \Delta\mu =\Delta\mu(l)-\Delta\mu(s) =-S_m\DeltaT

[\mu(l-5°C)-\mu(l,0°C)] =  [\mu(s-5°C)-\mu(s,0°C)]=-S_mΔT

\mu(l,-5°C)-\mu(s,-5°C)=-Sm\DeltaT [\mu(l,0

\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)

     = 109J/mole

6 0
2 years ago
What is the part of the cartridge that actually flies outward (the projectile)?
zhannawk [14.2K]

Answer:

the culotte, the fulminant capsule and the sheath or cap together with remains of powder

Explanation:

A bullet has a nose, sheath or casing, a short propellant charge, and a capsule.

All these components are part of the vouchers. The classification of these is in metallic and semi-metallic.

3 0
2 years ago
The specific heat capacity of gold is 0.128 J/g C. How much would be needed to warm 250.0 grams of gold from 25.0 C to 100.0 C.
madam [21]

Answer:

3,200 joules

Explanation:

q = mcΔT = (250.0 g)(0.128 j/g°C)(100°C -25°C) = 3,200 joules

7 0
3 years ago
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