Answer: -
1) 8.33 minutes
2) 118.39 in/ s
180.43 m/min
10.83 km/ hr
Explanation: -
Speed of light = 3 x 10⁸ m/s
Distance of the earth from the sun= 93 million miles
We know 1 million = 1,000,000
Also 1 mile = 1609 m
Distance of the earth from the sun= 93 million miles
= 93,000,000 miles.
= 1.5 x 
m
Time taken = 
= ![\frac{1.5 x [tex] 10^{11}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1.5%20x%20%5Btex%5D%2010%5E%7B11%7D%20%20)
m}{3 x 10⁸ m/s} [/tex]
= 500 s
= 500/ 60
= 8.33 minutes
2) Distance = 1 mile = 63360 inches
Time taken = 8.92 min
= 8.92 x 60
= 535.2 s
Speed = 
= 
= 118.39 in/ s
Distance = 1 mile = 63360 inches = 63360 x 2.54 cm = 63360 x 2.54 x 
m
Time taken = 8.92 min
Speed =
= ![\frac{63360 x 2.54 x [tex] 10^{-2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B63360%20x%202.54%20x%20%5Btex%5D%2010%5E%7B-2%7D%20%20)
m}{8.92 min} [/tex]
= 180.43 m/ min
1 m = 10⁻³ Km
1 min = 1/60 hour
1 m /min = 10⁻³ km/ 
= 60/1000
=0.06 km/hr
180.43 m / min = 180 x 0.06 km / hr
= 10.93 km / hr
Answer:
Explanation:
Hello there!
In this case, since these problems about gas mixtures are based off Dalton's law in terms of mole fraction, partial pressure and total pressure, we can write the following for hydrogen, we are given its partial pressure:
And can be solved for the total pressure as follows:
However, we first calculate the mole fraction of hydrogen by subtracting that of nitrogen to 1 due to:
Then, we can plug in to obtain the total pressure:
Regards!
<u>Answer:</u> The standard enthalpy change of the reaction is coming out to be -16.3 kJ
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H_f_%7B%28MgCl_2%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_f_%7B%28H_2O%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_f_%7B%28Mg%28OH%29_2%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_f_%7B%28HCl%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-641.8%29%29%2B%282%5Ctimes%20%28-241.8%29%29%5D-%5B%281%5Ctimes%20%28-924.5%29%29%2B%282%5Ctimes%20%28-92.30%29%29%5D%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D-16.3kJ)
Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ
Answer:
The answer to your question is letter C.
Explanation:
Reaction
Potassium hydroxide = KOH
Barium chloride = BaCl₂
Potassium chloride = KCl
Barium hydroxide = Ba(OH)₂
KOH + BaCl₂ ⇒ KCl + Ba(OH)₂
Reactant Elements Products
1 K 1
1 Ba 1
2 Cl 1
1 H 2
1 O 2
The reaction is unbalanced
2KOH + BaCl₂ ⇒ 2KCl + Ba(OH)₂
Reactant Elements Products
2 K 2
1 Ba 1
2 Cl 2
2 H 2
2 O 2
Now, the reaction is balanced
Answer:
Option B. Malleable, Conductor, High melting point, Lustrous
Explanation:
Mg has a higher melting point because of the strong electrostatic force of attraction between the magnesium ions (Mg^2+). The rest properties listed are all general properties of metals
