Answer:
319.15^{o}C[/tex]
Explanation:
When all other variables are constant, we are allowed to use the formula
![\frac{T_{2} }{V_{2} } = \frac{T_{1} }{V_{1} } \\Which can be rewritten as T_{2} = \frac{T_{1} V_{2} }{V_{1} }if you make T2 the subject of the formula. This formula is true only if temperature is in Kelvin not degrees Celsius so T1 must be converted to KelvinNow to calculate T2[tex]T_{2}= \frac{296.15K*3.38.10^{3}L }{1.69.10^{3}L }= 592.3K](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7B2%7D%20%7D%7BV_%7B2%7D%20%7D%20%3D%20%5Cfrac%7BT_%7B1%7D%20%7D%7BV_%7B1%7D%20%7D%20%5C%5C%3C%2Fp%3E%3Cp%3EWhich%20can%20be%20rewritten%20as%20T_%7B2%7D%20%3D%20%5Cfrac%7BT_%7B1%7D%20V_%7B2%7D%20%7D%7BV_%7B1%7D%20%7D%3C%2Fp%3E%3Cp%3Eif%20you%20make%20T2%20the%20subject%20of%20the%20formula.%20This%20formula%20is%20true%20only%20if%20temperature%20is%20in%20Kelvin%20not%20degrees%20Celsius%20so%20T1%20must%20be%20converted%20to%20Kelvin%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3ENow%20to%20calculate%20T2%3C%2Fp%3E%3Cp%3E%5Btex%5DT_%7B2%7D%3D%20%5Cfrac%7B296.15K%2A3.38.10%5E%7B3%7DL%20%7D%7B1.69.10%5E%7B3%7DL%20%7D%3D%20592.3K)
= 
Answer:
4.21 g of AgCl
3.06 g of BaCl₂ will be needed to complete the reaction
Explanation:
The first step is to determine the reaction.
Reactants: BaCl₂ and AgNO₃
The products will be the silver chloride (AgCl) and the Ba(NO₃)₂
The reaction is: BaCl₂(aq) + 2AgNO₃(aq) → 2AgCl(s) ↓ + Ba(NO₃)₂ (aq)
We determine the silver nitrate moles: 5 g . 1mol / 169.87 g = 0.0294 moles. Now, according to stoichiometry, we know that ratio is 2:2-
2 moles of nitrate can produce 2 moles of chloride, so the 0.0294 moles of silver nitrate, will produce the same amount of chloride.
We convert the moles to mass → 143.32 g / mol . 0.0294 mol = 4.21 g of AgCl.
Now, we consider the BaCl₂.
2 moles of nitrate can react to 1 mol of barium chloride
Then, 0.0294 moles of silver nitrate will react to (0.0294 . 1) /2 = 0.0147 moles. We convert the moles to mass:
0.0147 mol . 208.23 g /1mol = 3.06 g of BaCl₂
Answer:
They are already matched for you. It goes
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2.
3.
in the order of the questions.
Answer:
696g
Explanation:
To determine the mass of ethylbutanoate produced, it is important that we know the number of moles of ethylbutanoate produced.
Now from the chemical equation, it can be seen that one mole of ethanol produces one mole of ethylbutanoate. This means that 6.0 mol of ethanol will surely produce 6.0 mol of ethylbutanoate.
Now, we need to get the mass of ethylbutanoate produced. The mass of ethylbutanoate produced is the number of moles of ethylbutanoate produced multiplied by the molar mass of ethylbutanoate produced.
The molar mass of ethylbutanoate is 12(6) + 12(1) + 16(2) = 116g/mol
The mass thus produced is 116 * 6 = 696g
Answer: 2369.84c
Explanation:
The length of the wire is 2369.84cmData Given;density = 8.94 g/cm^3diameter = 9.50mm = 0.95cm radius = d / 2 = 0.95/2 = 0.475cmmass (not given, but assuming m = 15kg = 15000kgDensity of the WireLet's calculate the volume of the wire using formula of density.substitute the values and solve for volumeLength of the WireTo calculate the length of the wire, we would use the volume of a cylinder on this.From the calculation above, the length of the copper is 2369.84c
