Well your mass wont change...so it would be 35 grams of a liquid
Answer:
The <u>equilibrium constant</u> is:

Explanation:
The correct equation is:
Thus, with the equilibrium concentrations you can calculate the equilibrium constant, Kc.
The equation for the equilibrium constant is:
![k_c=\dfrac{[NH_3]^2}{[N_2]\cdot [H_2]^3}](https://tex.z-dn.net/?f=k_c%3D%5Cdfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5Ccdot%20%5BH_2%5D%5E3%7D)
Substituting:


Answer:
5.0 × 10²⁴ molecules
Explanation:
Step 1: Write the balanced double displacement reaction
2 NaOH + CuSO₄ ⇒ Na₂SO₄ + Cu(OH)₂
Step 2: Calculate the moles corresponding to 5.0 × 10²⁴ molecules of Na₂SO₄
We will use Avogadro's number: there are 6.02 × 10²³ molecules in 1 mole of molecules.
5.0 × 10²⁴ molecule × 1 mol/6.02 × 10²³ molecule = 8.3 mol
Step 3: Calculate the moles of CuSO₄ required to produce 8.3 moles of Na₂SO₄
The molar ratio of CuSO₄ to Na₂SO₄ is 1:1. The moles of CuSO₄ required are 1/1 × 8.3 mol = 8.3 mol.
Step 4: Calculate the molecules corresponding to 8.3 moles of CuSO₄
We will use Avogadro's number.
8.3 mol × 6.02 × 10²³ molecule/1 mol = 5.0 × 10²⁴ molecule
<span>(2.09 mL) x (1.592 g/mL) / (227.0871 g C3H5O9N3/mol) = 0.014652 mole C3H5O9N
4 moles C3H5O9N produce 12 + 6 + 1 + 10 = 29 moles of gases, so:
(0.014652 mole C3H5O9N) x (29/4) = 0.106 mole of gases
(b)
(0.106 mol) x (46 L/mol) = 4.88 L gases
(c)
(0.014652 mole C3H5O9N) x (6/4) x (28.0134 g/mol) = 0.616 g N2</span>