Its a. <span>an oceanic-continental convergent boundary</span>
Answer:
0.5ppm
Explanation:
Step 1:
Data obtained from the question.
Volume of water = 2500L
Mas of Cu = 1.25 g
Step 2:
Determination of the concentration of Cu in g/L. This is illustrated below:
Volume of water = 2500L
Mas of Cu = 1.25 g
Conc. of Cu In g/L =?
Conc. g/L = Mass /volume
Conc. of Cu in g/L = 1.25/2500
Conc. of Cu in g/L = 5x10^–4 g/L
Step 3:
Conversion of the concentration of Cu in g/L to ppm. This is illustrated below
Recall:
1g/L = 1000mg/L
Therefore, 5x10^–4 g/L = 5x10^–4 x 1000 = 0.5mg/L
Now, we know that 1mg/L is equal to 1ppm.
Therefore, 0.5mg/L is equivalent to 0.5ppm
Answer:
Explanation:
Mitochondria are a part of eukaryotic cells. The main job of mitochondria is to perform cellular respiration. This means it takes in nutrients from the cell, breaks it down, and turns it into energy. This energy is then in turn used by the cell to carry out various functions.
Answer:
378mL
Explanation:
The following data were obtained from the question:
Pressure (P) = 99.19 kPa
Temperature (T) = 28°C
Number of mole (n) = 0.015 mole
Volume (V) =...?
Next, we shall convert the pressure and temperature to appropriate units. This is illustrated below:
For Pressure:
101.325 KPa = 1 atm
Therefore, 99.19 kPa = 99.19/101.325 = 0.98 atm
For Temperature:
T(K) = T(°C) + 273
T(°C) = 28°C
T(K) = 28°C + 273 = 301K.
Next we shall determine the volume of N2. The volume of N2 can be obtained by using the ideal gas equation as shown below:
PV = nRT
Pressure (P) = 0.98 atm
Temperature (T) = 301K
Number of mole (n) = 0.015 mole
Gas constant (R) = 0.0821atm.L/Kmol.
Volume (V) =...?
0.98 x V = 0.015 x 0.0821 x 301
Divide both side by 0.98
V = (0.015 x 0.0821 x 301) /0.98
V = 0.378 L
Finally, we shall convert 0.378 L to millilitres (mL). This is illustrated below:
1L = 1000mL
Therefore, 0.378L = 0.378 x 1000 = 378mL
Therefore, the volume of N2 collected is 378mL
Answer:
a?
Explanation:cause if organism w is a fossil and organism y is a rock the rock is way older than a fossil so a
