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egoroff_w [7]
4 years ago
13

How did scientists carry out research before and after the scientific revolution?

Chemistry
1 answer:
iren [92.7K]4 years ago
6 0
They make scientific hypothiesis
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The monomer of poly(vinyl chloride) has the formula C2H3Cl. If there are 1,565 repeat units in a single chain of the polymer, wh
monitta

Answer:

\large \boxed{9.780 \times 10^{4}\text{ u}}

Explanation:

The molecular mass of a monomer unit is:

C₂H₃Cl = 2×12.01 + 3×1.008 + 35.45 = 24.02 + 3.024 + 35.45 = 62.494 u

For 1565 units,

\text{Molecular mass} = \text{1565 units} \times \dfrac{\text{62.494 u}}{\text{1 unit }} = \mathbf{9.780 \times 10^{4}}\textbf{ u}\\\\\text{The molecular mass of the chain is $\large \boxed{\mathbf{9.780 \times 10^{4}}\textbf{ u}}$}

8 0
3 years ago
Calculate δg o for each reaction using δg of values:(a) h2(g) + i2(s) → 2hi(g) kj (b) mno2(s) + 2co(g) → mn(s) + 2co2(g) kj (c)
steposvetlana [31]
Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol

Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol

Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
      = (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
      = 176.15 kJ - 84.78 kJ = 91.38 kJ 
 





 
6 0
3 years ago
Read 2 more answers
What is the proper name for the common sunflower
harkovskaia [24]

Answer: Helianthus L. Sunflower

Explanation:

6 0
3 years ago
Read 2 more answers
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔
Vilka [71]

Answer : The correct option is, (C) 1.1

Solution :  Given,

Initial moles of N_2O_4 = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

The given equilibrium reaction is,

                           N_2O_4(g)\rightleftharpoons 2NO_2(g)

Initially                      c                 0

At equilibrium   (c-c\alpha)           2c\alpha

The expression of K_c will be,

K_c=\frac{[NO_2]^2}{[N_2O_4]}

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

where,

\alpha = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}

K_c=1.066\aprrox 1.1

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0
3 years ago
Which one of these is NOT a sign of chemical change? *
Gekata [30.6K]
I think it’s light
Have fun :)
3 0
3 years ago
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