Answer:
1-b
2-weaker(option is incorrect)
3-a
Explanation:
1-b because iodine is more electronegative because of this negative on iodine will be more stable as negative charge on more electronegative element is more stable.
2-weaker as size of Te (Tellurium) is greater than S (sulphur) so bond length of H-Te is larger than H-S and therefore bond energy will be lesser and can easily give hydrogen in case of H-Te. as bond energy is inversly proportional to bond length.
3-a hydrogen has more negative electron affinity so hydrogen will have -1 charge and it will behave as a electron donar atom that is basic not acidic hence NaH is not acidic.
Answer:
0.21 g
Explanation:
The equation of the reaction is;
NaCl(aq) + AgNO3(aq) -----> NaNO3(aq) + AgCl(s)
Number of moles of NaCl= 0.0860 g /58.5 g/mol = 0.00147 moles
Number of moles of AgNO3 = 30/1000 L × 0.050 M = 0.0015 moles
Since the reaction is 1:1, NaCl is the limiting reactant.
1 mole of NaCl yields 1 mole of AgCl
0.00147 moles of NaCl yields 0.00147 moles of AgCl
Mass of precipitate formed = 0.00147 moles of AgCl × 143.32 g/mol
= 0.21 g
I believe the correct answer from the choices listed above is the first option. The compound that contains both ionic and covalent bonding is KOH or potassium hydroxide. It contains one covalent<span> (O-H) and one that is </span>ionic<span> (K-O). Hope this helps.</span>
Answer:
The answer is 0.36 kg/s NO
Explanation:
the chemical reaction of NH3 to NO is as follows:
4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l)
We have the following data:
O2 Volume rate = 645 L/s
P = 0.88 atm
T = 195°C + 273 = 468 K
NO molecular weight = 30.01 g/mol
we calculate the moles found in 645 L of O2:
P*V = n*R*T
n = P*V/R*T
n= (0.88 atm * 645L/s)/((0.08205 L*atm/K*mol) * 468 K) = 14.78 moles of O2
With the reaction we can calculate the number of moles of NO and with its molecular weight we will have the rate of NO:
14.78 moles/s O2 * 4 molesNO/5 molesO2 * 30.01 g NO/1 molNO x 1 kgNO/1000 gNO = 0.36 kg/s NO
Hey there!
Atomic mass Bromine ( Br ) = 79.9 u
Therefore:
79.9 g Br --------------- 22.4 L ( at STP )
18.0 g Br --------------- volume ??
Volume Br = 18.0 * 22.4 / 79.9
Volume Br = 403.2 / 79.9
Volume Br = 5.046 L
hope this helps!