Answer:
The symbol for decimeter is dm
Explanation:
No need for an explanation
1°/ . 2 Al + 6 HCl → 2 AlCl3 + 3 H2
<span>k1 = n(Al) / 2 = 4,5 / 2 = 2,25 </span>
<span>k2 = n(HCl) / 6 = 11,5 / 6= 1,92 </span>
<span>k2 < k1 ==> HCl is the limiting reactant </span>
<span>6 mol of HCl ---> 2 mol of H2 </span>
<span>11,5 mol of HCl ---> 3,83 mol of H2 </span>
The formula for solving current given with resistance and power source or voltage is shown below:
I = V/R
When two 5 ohms resistors are in series, we have:
I = 9 volts / (5+5 ohms)
I = 0.9 amperes
When it is being added with another 7.5 resistors, we have:
I = 9 volts / (5+5+7.5 ohms)
I = 0.529 ampere
The answer to the question is the letter "D. decrease; 0.51 amps".
Answer:
Explanation:
Hello.
In this case, taking into account that HCl has one molecule of hydrogen per mole of compound which weights 36.45 g/mol, we compute the number of molecules of hydrogen in hydrochloric acid by considering the given mass and the Avogadro's number:
Now, from the 180 g of water, we see two hydrogen molecules per molecule of water, thus, by also using the Avogadro's number we compute the molecules of hydrogen in water:
Thus, the total number of molecules turns out:
Regards.
Answer:
222.30 L
Explanation:
We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:
Mass of NH₃ = 100 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 100 / 17
Mole of NH₃ = 5.88 moles
Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
3 moles of H₂ reacted to produce 2 moles NH₃.
Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e
Xmol of H₂ = (3 × 5.88)/2
Xmol of H₂ = 8.82 moles
Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:
Pressure (P) = 95 KPa
Temperature (T) = 15 °C = 15 + 273 = 288 K
Number of mole of H₂ (n) = 8.82 moles
Gas constant (R) = 8.314 KPa.L/Kmol
Volume (V) =?
PV = nRT
95 × V = 8.82 × 8.314 × 288
95 × V = 21118.89024
Divide both side by 95
V = 21118.89024 / 95
V = 222.30 L
Thus the volume of Hydrogen needed for the reaction is 222.30 L
