Como son los recorridos de los oleoductos y gasoductos en argentina?

What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP?

BigorU [14]

Mg reaction with O₂ gas will produce MgO so the equation will be
2Mg+O₂⇒2MgO. (You have to find the equation in order two figure out the number of moles of O₂ that will react with 1 mole of MgO).

The first step is to find the number of moles of Mg in 4.03g of Mg. You can do this by dividing 4.03g Mg by its molar mass (which is 24.3g/mol) to get 0.1658mol Mg. Then you have to find the number of moles of O₂ that will react with 0.1658mol Mg. To do this you need to use the fact that 1mol O₂ will react with 2mol Mg (this reatio is from the chemical equation) so you have to multiply 0.1658mol Mg by (1mol O₂)/(2mol Mg) to get 0.0829mol O₂. From here you would usually use PV=nRT and solve for V However, the question tells us that we are at STP, that means you can use the fact that 22.4L of gas is 1 mol of gas at STP. Using that information we can find the volume of O₂ gas by mulitlying 0.0829mol O₂ by 22.4L/mol to get 1.857L which equals 1857mL.
therefore, 1857mL of O₂ gas will react with 4.03g of Mg.

I hope this helps. Let me know in the comments if anything is unclear.

6 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

You mix two clear liquids and a colored solid forms at the bottom of the beaker. Do you think this is a chemical change? How can

UkoKoshka [18]

The answer is yes. A chemical change occurred.
Chemical change is defined as the rearrangement or alteration in the of atoms in one or more substance which result in the formation of a new substance.

In the above, you mixed two clear liquids and the result was a new substance which is a colored solid precipitate at the bottom of the beaker.
This means that changes in the atoms of the two clear liquids occurred leading to the formation of this new solid substance.

This means that chemical change has occurred.

Hope this helps :)

7 0

3 years ago

Which units are used to measure both velocity and speed? Select three options

Ede4ka [16]

Answer and Explanation:

The options aren't listed in your question, but here are some units that are regularly and normally used (in the classroom and in the outside world):

(The SI unit of distance and displacement is the meter. The SI unit of time is the second.)

Meters per Second (m/s)

kilometers per hour (km/hr)

kilometers per second (km/sec)

To find the average speed, you do distance divided by time.

To find the average velocity, you do the final position minus the initial position, divided by the final time minus the initial time.

#teamtrees #PAW (Plant And Water)

I hope this helps!

8 0

3 years ago

Explain why the process of dissolution can be exothermic or endothermic (heat of solution) with different solutes

RUDIKE [14]

Answer:

The process of dissolving is exothermic when more energy is released when water molecules “bond” to the solute than is used to pull the solute apart. Because more energy is released than is used, the molecules of the solution move faster, making the temperature increase.

Project the image Endothermic Dissolving.

The process of dissolving is endothermic when less energy is released when water molecules “bond” to the solute than is used to pull the solute apart. Because less energy is released than is used, the molecules of the solution move more slowly, making the temperature decrease.

4 0

2 years ago