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Vladimir79 [104]
3 years ago
5

1. How much magnesium oxide is produced if 12.1 g of magnesium reacts with 16 g of oxygen

Chemistry
1 answer:
melamori03 [73]3 years ago
8 0

Answer:

20.2g (3 s.f.)

Explanation:

Please see attached picture for full solution.

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How many grams of sodium oxide can be produced when 55.3 g Na react with 64.3 g O2?. . Unbalanced equation: Na + O2 → Na2O. . Sh
GenaCL600 [577]
 <span>this is a limiting reagent problem. 

first, balance the equation 
4Na+ O2 ---> 2Na2O 

use both the mass of Na and mass of O2 to figure out how much possible Na2O you could make. 
start with Na and go to grams of Na2O 

55.3 gNa x (1molNa/23.0gNa) x (2 molNa2O/4 molNa) x (62.0gNa2O/1molNa2O) = 75.5 gNa2O 

do the same with O2 

64.3 gO2 x (1 molO2/32.0gO2) x (2 molNa2O/1 mol O2) x (62.0gNa2O/1molNa2O) = 249.2 g Na2O 

now you must pick the least amount of Na2O for the one that you actually get in the reaction. This is because you have to have both reacts still present for a reaction to occur. So after the Na runs out when it makes 75.5 gNa2O with O2, the reaction stops. 

So, the mass of sodium oxide is 

75.5 g</span>
3 0
3 years ago
Read 2 more answers
If a helium balloon is put into a hot oven. What happens to the volume of the balloon?
AveGali [126]

Answer:

What about hot air balloons? They work by similar principles. If you heat up a gas it expands. In the case of a hot air balloon, when the gas inside the balloon expands the extra gas is pushed out the bottom of the balloon, meaning that there are fewer atoms inside the balloon, meaning that the air in the balloon is lighter than the air outside the balloon.

The amount of lifting power is controlled by how hot the air is. If you heat the air inside the balloon 100 degrees F hotter than the outside air temperature, then the air inside the balloon will be about 25 percent lighter than the air outside the balloon. So a cubic foot of air weighs about 35 grams at 32 degrees F. A cubic foot of hot air at 132 degrees F will weigh 25 percent less, or about 26.5 grams. The difference is 8.5 grams or so. So a hot air balloon has to be much bigger to support the same weight, but it will float because hotter air is lighter than cooler air.

Explanation:

The amount of lifting power

6 0
3 years ago
The period of a ripple on a lake is half a second. what is the frequency of the wave
Nataly_w [17]

Answer:

2 Hertz

Explanation:

<em>The frequency would be 2 Hertz.</em>

<u>The frequency of a wave is defined as the rate at which the particles of a medium vibrates when the wave is passed through it while the period of a wave is the time it takes the particles to make a complete cycle of vibration.</u>

The frequency of a wave is inversely related to its period and is defined by the following equation:

f = 1/t, where f is the frequency (in hertz) and t is the period (in seconds).

Hence, if the period of a ripple is 1/2 or 0.5 seconds, the frequency becomes;

f = 1/0.5 = 2 Hertz

7 0
3 years ago
PH of 0,035M HCl? How to count if your calculator hasn't got -lg?
IrinaK [193]

<u>Given information:</u>

Concentration of HCl = 0.035 M

<u>To determine:</u>

pH of the solution

<u>Explanation:</u>

Hydrochloric acid, HCl is a strong acid. It will completely dissociate to give H+ and Cl- ions

HCl → H+ + Cl-

Hence the concentration of H+ = Cl- = 0.035M

Now, pH measures the strength of H+ ions in a given solution. It is expressed as:

pH = -log[H+]

pH (HCl) = -log(0.035) = 1.46

Ans: pH of 0.035M HCl is 1.46

6 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
3 years ago
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