Answer:
2. precipitate is formed
Explanation:
In this reaction barium nitrate and sodium sulfate solutions react to form a barium sulfate precipitate.Barium and sulfate ions will react to give barium sulfate precipitate where as the sodium and nitrate ions are spectator ions.
The net ionic equation after removing unchanged ions from each side of the full ionic equation will be;
Ba²⁺+ SO₄²⁻ → BaSO₄ (s)
Answer:
10.54 L
Explanation:
Given that:-
Moles of = 1.00 moles
According to the given reaction:-
7 moles of oxygen gas reacts with 4 moles of ammonia
1 mole of oxygen gas reacts with moles of ammonia
Moles of ammonia = 0.5714 moles
Also, Given that:
Temperature = 850 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (850 + 273.15) K = 1123.15 K
n = 0.5714 moles
P = 5.00 atm
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L atm/ K mol
Applying the equation as:
5.00 atm × V = 0.5714 moles ×0.0821 L atm/ K mol × 1123.15 K
⇒V = 10.54 L
<u>10.54 L of required.</u>
Part A: Given mass of NaCl * 1mol of NaCl/molar mass of NaCl = # of moles
17.45 g NaCl * 1 mole of NaCl /58.44 g NaCl = 0<em>.298 moles of NaCl</em>
Part B: First, let’s start with sodium, Na. To find the given mass of sodium, divide the given mass of NaCl (17.45)/the molar mass of NaCl(58.44) and multiply that by the molar mass of Na(22.99) to get the given mass of Na.
Given mass of Na * 1mol of Na/molar mass of Na= # of moles of Na.
6.864* 1mol Na/22.99 g Na=0.298 moles of sodium.
Repeat the above steps with chlorine.
Given mass of Na or Cl*1mol of Cl/molar mass of Cl = # of moles Cl.
10.58* 1mol Cl/ 35.45 g of Cl= <em>0.298 moles of chlorine.</em>
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Good Luck :)
Btw sry for answering so late.
Carbon can form four covalent bonds. Covalent bonds are chemical bonds that form between nonmetals. In a covalent bond, two atoms share a pair of electrons. By forming four covalent bonds, carbon shares four pairs of electrons, thus filling its outer energy level and achieving stability.