Answer:
a) 320: two significant figures.
b) 2,366: four significant figures.
c) 73.0: three significant figures.
d. 532.5: four significant figures.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to write each number by knowing we move the decimal places to the right as much as the exponent is, and also, we count every figure, even zeros, because they are to the right of the first nonzero digit:
a) 320: two significant figures because the rightmost zero is not preceded o followed by a decimal place.
b) 2,366: four significant figures.
c) 73.0: three significant figures, because the zero is followed by the decimal place.
d. 532.5: four significant figures.
Regards!
Answer:
hey just where are the rections keep a pic
Explanation:
okay?.........................................
Answer:
0.184 atm
Explanation:
The ideal gas equation is:
PV = nRT
Where<em> P</em> is the pressure, <em>V</em> is the volume, <em>n</em> is the number of moles, <em>R</em> the constant of the gases, and <em>T</em> the temperature.
So, the sample of N₂O₃ will only have its temperature doubled, with the same volume and the same number of moles. Temperature and pressure are directly related, so if one increases the other also increases, then the pressure must double to 0.092 atm.
The decomposition occurs:
N₂O₃(g) ⇄ NO₂(g) + NO(g)
So, 1 mol of N₂O₃ will produce 2 moles of the products (1 of each), the <em>n </em>will double. The volume and the temperature are now constants, and the pressure is directly proportional to the number of moles, so the pressure will double to 0.184 atm.
Answer:
1.667L of a 0.30M BaCl₂ solution
Explanation:
<em>Of a 0.30M barium chloride, contains 500.0mmol of barium chloride.</em>
<em />
Molarity is an unit of concentration used in chemistry defined as the moles of solute present in 1 liter of solution.
In a 0.30M BaCl₂ solution there are 0.30 moles of BaCl₂ in 1 liter of solution.
Now, in your solution you have 500mmol of BaCl₂ = 0.500 moles of BaCl₂ (1000 mmol = 1 mol). Thus, 0.500 moles of BaCl₂ are present in:
0.500 moles * (1L / 0.30 moles) =
<h3>1.667L of a 0.30M BaCl₂ solution</h3>
