Answer: There are 0.5 grams of barium sulfate are present in 250 of 2.0 M 
solution.
Explanation:
Given: Molarity of solution = 2.0 M
Volume of solution = 250 mL
Convert mL int L as follows.
Molarity is the number of moles of solute present in liter of solution. Hence, molarity of the given solution is as follows.
Thus, we can conclude that there are 0.5 grams of barium sulfate are present in 250 of 2.0 M solution.
Answer:
It works because of the significant principle of piezoelectricity
Explanation:
Mass Molar of
Ca = 3*40 = 120 amu
P = 2*31= 62 amu
O = (16*4)*2 = 64*2 = 128 amu
--------------------------------------
Mass Molar of
= 120 + 62 + 128 = 310 g/mol
Therefore: <span>What is the gram formula mass of Ca3(PO4)2 ?
</span>Answer:
310 grams
Answer:
all th eabove i had this pls give brainliest i got this correct of of edmentum
Explanation:
Answer:
![[base]=0.28M](https://tex.z-dn.net/?f=%5Bbase%5D%3D0.28M)
Explanation:
Hello,
In this case, by using the Henderson-Hasselbach equation one can compute the concentration of acetate, which acts as the base, as shown below:
![pH=pKa+log(\frac{[base]}{[acid]} )\\\\\frac{[base]}{[acid]}=10^{pH-pKa}\\\\\frac{[base]}{[acid]}=10^{4.9-4.76}\\\\\frac{[base]}{[acid]}=1.38\\\\](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D10%5E%7BpH-pKa%7D%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D10%5E%7B4.9-4.76%7D%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D1.38%5C%5C%5C%5C)
![[base]=1.38[acid]=1.38*0.20M=0.28M](https://tex.z-dn.net/?f=%5Bbase%5D%3D1.38%5Bacid%5D%3D1.38%2A0.20M%3D0.28M)
Regards.
