This is an example of Charles’ Law problems, the basic equation is: V1/T1 = V2/T2. One vital thing to recall for all gas law problems is that the temperature must be in Kelvin (not Celsius).
So our given is 10.0 C = 283 K. So
V1/T1 = V2/T2
733/283 = 950/T2
T2 = 367 K
Answer:
1.71 × 10²¹ molecules
Explanation:
1 gram is equal to 1000 milligrams. The mass in grams corresponding to 500 mg is:
500 mg × (1 g / 1000 mg) = 0.500 g
The molar mass of ascorbic acid is 176.12 g/mol. The moles corresponding to 0.500 grams of ascorbic acid are:
0.500 g × (1 mol/ 176.12 g) = 0.00284 mol
In 1 mole of ascorbic acid, there are 6.02 × 10²³ molecules of ascorbic acid (Avogadro's number). The molecules in 0.00284 moles are:
0.00284 mol × (6.02 × 10²³ molecules/ 1 mol) = 1.71 × 10²¹ molecule
Answer:
Enzymes may require a nonprotein cofactor or ion for catalysis to take speed up more appreciably than if the enzymes act alone;
Enzymes increase the rate of chemical reaction by lowering activation energy barriers.
Explanation:
Some enzymes need a cofactor to act, it is attached to the enzyme and can be nonprotein such as a metal ion. The enzyme function depends on the physical properties of the environmental, especially temperature and pH, each enzyme has a great point of pH and temperature where it has a maximum activity.
If the three-dimensional function of an enzyme is altered, it loses it specified and may not catalyze the reaction, because the structure of the enzyme is responsable for its specified. The catalyst occurs because the enzyme lows the activation energy barriers and this increases the rate of the reaction.
Climate change or it could just be Climate,
Hopefully that helps ❤
The chalk particles embed themselves into the small pores on the surface.
Although a chalkboard seems smooth to the touch, it is quite rough at the microscopic level, with <em>pores</em> that reach below the surface.
When you drag chalk across the board, friction causes small particles of chalk to rub off onto the surface.
If you leave the markings for a long time, some of the chalk particles will work their way into the pores.
A brush will remove the surface particles, but <em>it will not be able to get at the particles in the pores</em>.
