The best answer is letter A.
The main factors that change the speed of enzymatic reactions are temperature, pH and substrate concentration (quantity). <span>There are enzymes that need other associated molecules to work. These molecules are called enzyme co-factors. They can be organic ions like mineral salts.</span>
Answer: -
Molality of NaCl = 2.807 ×10⁻² m
Explanation: -
Molarity given = 2.800×10⁻² M
This means there are 2.800×10⁻² moles of NaCl per 1000 mL of solution.
Volume of water = 999.2 ml
Density of water = 0.9982 g/ml
Mass of water = Density of water x Volume of water
= 0.9982 g/ml x 999.2 ml
= 997.4 g
Thus 997.4 g of water has 2.800×10⁻² moles of NaCl.
1000 g of water has x 2.800×10⁻² moles of NaCl.
= 2.807 ×10⁻² moles of NaCl.
Molality of NaCl = 2.807 ×10⁻² m
<h3>
Answer:</h3>
Empirical formula is CrO
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of sample of Chromium as 7.337 gram
- Mass of the metal oxide formed as 9.595 g
We are required to determine the empirical formula of the metal oxide.
<h3>Step 1 ; Determine the mass of oxygen used </h3>
Mass of oxygen = Mass of the metal oxide - mass of the metal
= 9.595 g - 7.337 g
= 2.258 g
<h3>Step 2: Determine the moles of chromium and oxygen</h3>
Moles of chromium metal
Molar mass of chromium = 51.996 g/mol
Moles of Chromium = 7.337 g ÷ 51.996 g/mol
= 0.141 moles
Moles of oxygen
Molar mass of oxygen = 16.0 g/mol
Moles of Oxygen = 2.258 g ÷ 16.0 g/mol
= 0.141 moles
<h3>Step 3: Determine the simplest mole number ratio of Chromium to Oxygen</h3>
Mole ratio of Chromium to Oxygen
Cr : O
0.141 mol : 0.141 mol
1 : 1
Empirical formula is the simplest whole number ratio of elements in a compound.
Thus the empirical formula of the metal oxide is CrO
Answer:
14
73%
Explanation:
The mean Number of years worked :
. (sum of service years) / employees in the
(8+13+15+3+13+28+4+12+4+26+29+3+10+3+17+13+15+15+23+13+12+1+14+14+17+16+7+27+18+24) /
(417 / 30)
= 13.9 years
= 14 years
The percentage of employees who have worked for atleast 10 years :
Number of employees with service years ≥ 10 years = 22 employees
Total number of employees
Percentage (%) = (22 / 30= * 100% = 0.7333 * 100% = 73.33% = 73%