Answer : The volume of the cube submerged in the liquid is, 29.8 mL
Explanation :
First we have to determine the mass of ice.
Formula used :
Given:
Density of ice = 
Volume of ice = 45.0 mL
The cube will float when 40.5 g of liquid is displaced.
Now we have to determine the volume of the cube is submerged in the liquid.
Thus, the volume of the cube submerged in the liquid is, 29.8 mL
The temperature reached after four hours of cooling is 140 F (or) 42 C.
Explanation:
A food worker cooled a cup of soup for two hours,
step 1: The temperature reached in <u>two hours</u>= 70 F (or) 21 C
step 2: Then for <u>four hours</u>, <u>it is twice the value</u>
∴ The temperature reached in four hours= 2(70)= 140 F (or) 2(21)= 42 C
Answer:
Use a ratio of 0.44 mol lactate to 1 mol of lactic acid
Explanation:
John could prepare a lactate buffer.
He can use the Henderson-Hasselbalch equation to find the acid/base ratio for the buffer.
![\text{pH} = \text{pK}_{\text{a}} + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\3.5 = 3.86 + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\\log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 3.5 - 3.86 = -0.36\\\\\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 10^{-0.36} = \mathbf{0.44}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%20%3D%20%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C3.5%20%3D%203.86%20%2B%20%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%20%3D%203.5%20-%203.86%20%3D%20-0.36%5C%5C%5C%5C%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%20%3D%2010%5E%7B-0.36%7D%20%3D%20%5Cmathbf%7B0.44%7D)
He should use a ratio of 0.44 mol lactate to 1 mol of lactic acid.
For example, he could mix equal volumes of 0.044 mol·L⁻¹ lactate and 0.1 mol·L⁻¹ lactic acid.
I would say D. Because if you displace too much water it would would either go to fast or not move at all... that's just my thought
Answer:
6.72M of HNO3
Explanation:
In the problem you are diluting the original HNO3 solution by the addition of some water. The final volume is:
290.7mL + 350.0mL = 640.7mL
And you are diluting the solution:
640.7mL / 350.0mL = 1.8306 times
As the original concentration was 12.3M, the final concentration will be:
12.3M / 1.8306 =
<h3>6.72M of HNO3</h3>
