Ask question

Ask question

All categories

3 years ago

9

5. 1.00 mol HNO3 is treated with 4.47 g of magnesium. Calculate the number of moles of

1 answer:

ruslelena [56]3 years ago

3 0

Answer:

The balanced equation is:

2 HNO3 + Mg ---> Mg(NO3)2 + H2

From the equation, we can see that we need twice the moles of HNO3 than the moles of Mg

Moles of Mg:

Molar mass of Mg = 24 g/mol

Moles = Given mass / Molar Mass

Moles of Mg = 4.47 / 24 = 0.18 moles (approx)

Hence, 2(moles of Mg) = 0.36 moles of HNO3 will be consumed

Number of moles of HNO3 after the reaction is finished is the number of unreacted moles of HNO3

Unreacted moles of HNO3 = Total Moles - Moles consumed

Unreacted moles of HNO3 = 0.64 moles (approx)

Since we approximated the value of moles of Mg, the value of remaining moles of HNO3 will also be approximate

From the given options, we can see that 0.632 moles is the closest value to our answer

Therefore, 0.632 moles will remain after the reaction

You might be interested in

The equation K + NaOH => Na + KOH should it react, true or false?

Andru [333]

Answer:

yes it should react yoyoyoyoyo

3 0

2 years ago

The change in velocity over a specific amount of time Includes speeding up. slowing down or changing direction. This is known as

Nana76 [90]

Answer:

Acceleration

Explanation:

Acceleration = $\frac{velocity}{time}$

3 0

2 years ago

What kind of bond is the result of the transfer of an electron?

MrRa [10]

An Ionic bond is the result of transfer of electrons between atoms

5 0

3 years ago

Read 2 more answers

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

$[\mu(l-5$ ° $C)-\mu(l,0$ ° $C)]$ = $[\mu(s-5$ ° $C)-\mu(s,0$ ° $C)]$ $=-S_m$ ΔT

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

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

= 109J/mole

6 0

2 years ago

Which of the following molecules has one lone pair of electrons?

BigorU [14]

You can automatically rule out CH₄ since it has no lone pairs at all around the central atom. Water has 2. Ammonia is the only Lewis structure that contains one lone pair.

7 0

2 years ago

Read 2 more answers