Answer:
B
Explanation:
bonding is a process of two different atoms sharing electrons for stability and these electrons are attracted by one atom losing it's electrons to another
Answer:
N2H4 + 2H2O2 ---->N2 + 4H2O
Explanation:
N=2 N=2
H=6 ->8 H=2 ->8
O=2 -> 4 O=1 -> 4
Add coefficients to hydrogen peroxide on the left and water on the right, so that there is an equal number of hydrogens and oxygens.
B because once the circuit is burned the lights will all go of
Answer:
116.5 g of SO₂ are formed
Explanation:
The reaction is:
S₈(g) + 8O₂(g) → 8SO₂ (g)
Let's identify the moles of sulfur vapor, by the Ideal Gases Law
We convert the 921.4°C to Absolute T° → 921.4°C + 273 = 1194.4 K
5.87 atm . 3.8L = n . 0.082 L.atm/mol.K . 1194.4K
(5.87 atm . 3.8L) / (0.082 L.atm/mol.K . 1194.4K) = n → 0.228 moles of S₈
Ratio is 1:8, 1 mol of sulfur vapor can produce 8 moles of dioxide
Then, 0.228 moles of S₈ must produce (0.228 . 8) /1 = 1.82 moles
We convert the moles to g → 1.82 moles . 64.06 g /1mol = 116.5 g
Answer:- 0.273 kg
Solution:- A double replacement reaction takes place. The balanced equation is:

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

= 63.8 g aluminum nitrate
From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.
We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

= 
sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate and then the grams are converted to kg.

= 0.273 kg
So, 0.273 kg of 35% m/m sodium chlorate solution are required.