Answer:
<em>The molarity of the solution is 0,47 M</em>
Explanation:
Molarity is a concentration measurement that expresses the moles of solute (in this case NaOH) in 1 liter of solution (1000ml). First we calculate the mass of 1 mol of NaOH, to calculate the moles in 58.8 g of said compound:
Weight 1 mol NaOH= Weight Na + Weight O + Weight H=23 g + 16 g +1 g
Weight 1 mol NaOH= 40 g/mol
40 g ---1 mol NaOH
58,8g---x= (58,8g x 1 mol NaOH)/40g =1,47 mol NaOH
3,1 L solution ------1,47 mol NaOH
1 L solution --------x= (1 L solution x 1,47 mol NaOH)/3,1 L solution
<em>x= 0,47 mol NaOH ---> The molarity of the solution is 0,47 M</em>
Answer:
a inorganic ion that aids the biochemical function of a protein a molecule that has a catalytic biochemical function on a substrate an organic molecule that assists with a protein's catalytic function a molecule that is biochemically altered during a catalytic reaction
Answer: 1.14
Explanation:
![HCl+NaOH\rightarrow NaCl+H_2O](https://tex.z-dn.net/?f=HCl%2BNaOH%5Crightarrow%20NaCl%2BH_2O)
To calculate the molarity of acid, we use the equation given by neutralization reaction:
![n_1M_1V_1=n_2M_2V_2](https://tex.z-dn.net/?f=n_1M_1V_1%3Dn_2M_2V_2)
where,
are the n-factor, molarity and volume of acid which is ![HCl](https://tex.z-dn.net/?f=HCl)
are the n-factor, molarity and volume of base which is NaOH.
We are given:
![n_1=1\\M_1=?\\V_1=10.0mL\\n_2=1\\M_2=0.1M\\V_2=7.2mL](https://tex.z-dn.net/?f=n_1%3D1%5C%5CM_1%3D%3F%5C%5CV_1%3D10.0mL%5C%5Cn_2%3D1%5C%5CM_2%3D0.1M%5C%5CV_2%3D7.2mL)
Putting values in above equation, we get:
![1\times M_1\times 10.0=1\times 0.1\times 7.2\\\\M_1=0.072M](https://tex.z-dn.net/?f=1%5Ctimes%20M_1%5Ctimes%2010.0%3D1%5Ctimes%200.1%5Ctimes%207.2%5C%5C%5C%5CM_1%3D0.072M)
To calculate pH of gastric juice:
molarity of
= 0.072
![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![pH=-log(0.072)=1.14](https://tex.z-dn.net/?f=pH%3D-log%280.072%29%3D1.14)
Thus the pH of the gastric juice is 1.14
There are four structural isomers with the molecular formula C₄H₉Br.
You start by writhing the linear carbon chain and the you start moving the bromide atom until you find all the compounds. In our case you have bromide atom in the position 1 and 2 on the carbon chain.
Then you branch the carbon chain and you have isobutane. Here the possible positions for bromide is 1 and 2 on the primary carbon chain.