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Ket [755]
3 years ago
9

Skeleton equation for

Chemistry
1 answer:
Verizon [17]3 years ago
3 0

Answer:

Magnesium oxide = MgO

Iron = Fe

Iron (III) Oxide = Fe₂O₃

Magnesium = Mg

Skeleton equation = MgO + Fe ⇒ Fe₂O₃ + Mg

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this is liquid water in the form of droplets that have condoned from atmospheric water vapor and than precipitated
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We want to find how much charge is on the electrons in a nickel coin. follow this method. a nickel coin has a mass of about 5 g.
Lady bird [3.3K]
The mass of a nickel coin is 5 g.
1 mol of Ni weighs 58 g. 1 mol contains 6.022 x 10²³ atoms of Ni.
therefore in 58 g  there are 6.022 x 10²³ atoms of Ni
then in 5 g the number of Ni atoms are - 6.022 x 10²³ /58 x 5 = 5.2 x 10²² Ni atoms
Therefore number of Ni atoms are 5.2 x 10²² atoms in a nickel coin

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3 years ago
What is this sign plz tell for 100 points
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How many moles of sucrose (C12H22O11) would be in 8.7 L of a 1.1 M solution of sucrose?
poizon [28]

Answer:

9.57 mol.

Explanation:

<em>Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.</em>

<em />

<em>M = (no. of moles of solute)/(V of the solution (L)).</em>

<em></em>

∴ M = (no. of moles of sucrose)/(V of the solution (L)).

1.1 M = (no. of moles of sucrose)/(8.7 L).

<em>∴ no. of moles of sucrose = (1.1 M)(8.7 L) = 9.57 mol.</em>

5 0
3 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
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