Answer:
1455.6
Explanation: you first convert 2250ml to l by dividing by 1000 so you get 2.25l then you use Boyles law which is p1v1=p2v2 then insert values
35.75*100=p2*2.25 then divide both sides by 2.25 then you get 1455.6
<h3>
Answer:</h3>
P₂ = 0.67 atm
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality<u>
</u>
<u>Chemistry</u>
<u>Gas Laws</u>
Boyle's Law: P₁V₁ = P₂V₂
- P₁ is pressure 1
- V₁ is volume 1
- P₂ is pressure 2
- V₂ is volume 2
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] P₁ = 2.02 atm
[Given] V₁ = 4.0 L
[Given] V₂ = 12.0 L
[Solve] P₂
<u>Step 2: Solve</u>
- Substitute in variables [Boyle's Law]: (2.02 atm)(4.0 L) = P₂(12.0 L)
- [Pressure] Multiply: 8.08 atm · L = P₂(12.0 L)
- [Pressure] [Division Property of Equality] Isolate unknown: 0.673333 atm = P₂
- [Pressure] Rewrite: P₂ = 0.673333 atm
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs as our smallest.</em>
0.673333 atm ≈ 0.67 atm
Answer:
A single molecule of water has been isolated for the first time by trapping it in a fullerene cage. Water molecules are never found alone — they are always hydrogen-bonded to other molecules of water or polar compounds.
While making small volumes of pure water in a lab is possible, it's not practical to “make” large volumes of water by mixing hydrogen and oxygen together. The reaction is expensive, releases lots of energy, and can cause really massive explosions.
While making small volumes of pure water in a lab is possible, it's not practical to “make” large volumes of water by mixing hydrogen and oxygen together. The reaction is expensive, releases lots of energy, and can cause really massive explosions.
A water molecule consists of three atoms; an oxygen atom and two hydrogen atoms, which are bond together like little magnets. The atoms consist of matter that has a nucleus in the centre. The difference between atoms is expressed by atomic numbers.
Explanation:
Answer:- 0.273 kg
Solution:- A double replacement reaction takes place. The balanced equation is:
![3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3](https://tex.z-dn.net/?f=3NaClO_3%2BAl%28NO_3%29_3%5Crightarrow%203NaNO_3%2BAl%28ClO_3%29_3)
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.
![0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})](https://tex.z-dn.net/?f=0.29L%28%5Cfrac%7B1000mL%7D%7B1L%7D%29%28%5Cfrac%7B22g%7D%7B100mL%7D%29)
= 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.
![63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})](https://tex.z-dn.net/?f=63.8gAl%28NO_3%29_3%28%5Cfrac%7B1mol%7D%7B212.99g%7D%29%28%5Cfrac%7B3molNaClO_3%7D%7B1molAl%28NO_3%29_3%7D%29%28%5Cfrac%7B106.44g%7D%7B1mol%7D%29)
= ![95.7gNaClO_3](https://tex.z-dn.net/?f=95.7gNaClO_3)
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.
![95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})](https://tex.z-dn.net/?f=95.7gNaClO_3%28%5Cfrac%7B100gSolution%7D%7B35gNaClO_3%7D%29%28%5Cfrac%7B1kg%7D%7B1000g%7D%29)
= 0.273 kg
So, 0.273 kg of 35% m/m sodium chlorate solution are required.