We can use the heat equation,
Q = mcΔT
where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).
Q = 11.2 kJ = 11200 J
m = <span>145 g
</span>c = ?
ΔT = (67 - 22) °C = 45 °C
By applying the formula,
11200 J = 145 g x c x 45 °C
c = 1.72 J g⁻¹ °C⁻¹
Hence, specific heat of benzene is 1.72 J g⁻¹ °C⁻¹.
Answer:
(molecular) 3 CaCl₂(aq) + 2 (NH₄)₃PO₄(aq) ⇄ Ca₃(PO₄)₂(s) + 6 NH₄Cl(aq)
(ionic) 3 Ca²⁺(aq) + 6 Cl⁻(aq) + 6 NH₄⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s) + 6 NH₄⁺(aq) + 6 Cl⁻(aq)
(net ionic) 3 Ca²⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s)
Explanation:
The molecular equation includes al the species in the molecular form.
3 CaCl₂(aq) + 2 (NH₄)₃PO₄(aq) ⇄ Ca₃(PO₄)₂(s) + 6 NH₄Cl(aq)
The ionic equation includes all the ions (species that dissociate in water) and the species that do not dissociate in water.
3 Ca²⁺(aq) + 6 Cl⁻(aq) + 6 NH₄⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s) + 6 NH₄⁺(aq) + 6 Cl⁻(aq)
The net ionic equation includes only the ions that participate in the reaction and the species that do not dissociate in water. In does not include <em>spectator ions</em>.
3 Ca²⁺(aq) + 2 PO₄³⁻(aq) ⇄ Ca₃(PO₄)₂(s)
Answer:
4
Explanation:
Relationship between wavenumber and Rydberg constant (R) is as follows:
Here, Z is atomic number.
R=109677 cm^-1
Wavenumber is related with wavelength as follows:
wavenumber = 1/wavelength
wavelength = 253.4 nm
Z fro Be = 4
Therefore, the principal quantum number corresponding to the given emission is 4.
If 0.25mg of atropine is in 1mL
so
0.50mg of atropine is in x
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