Answer:
Explanation:
Every A will be paired with a T, and every C with a G and vice versa (T to A, and G to C)
So the first letters in the sequence are ATGGG. So the pair will be TACCC.
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:
The given molecules are SO2 and BrF5.
Explanation:
Consider the molecule SO2:
The central atom is S.
The number of domains on S in this molecule is three.
Domain geometry is trigonal planar.
But there is a lone pair on the central atom.
So, according to VSEPR theory,
the molecular geometry becomes bent or V-shape.
Hybridization on the central atom is
.
Consider the molecule BrF5:
The central atom is Br.
The number of domains on the central atom is six.
Domain geometry is octahedral.
But the central atom has a lone pair of electrons.
So, the molecular geometry becomes square pyramidal.
The hybridization of the central atom is 
.
The shapes of SO2 and BrF5 are shown below:
First, we have to get:
1- The heat required to increase T of ice from -50 to 0 °C:
according to q formula:
q1 = m*C*ΔT
when m is the mass of ice = mol * molar mass
= 1 mol * 18 mol/g
= 18 g
and C is the specific heat capacity of ice = 2.09 J/g-K
and ΔT change in temperature = 0- (-50) = 50°C
by substitution:
∴q1 = 18 g * 2.09 J/g-K *50°C
= 1881 J = 1.881 KJ
2- the heat required to melt this mass of ice is :
q2 = n*ΔHfus
when n is the number of moles of ice = 1 mol
and ΔHfus = 6.01 KJ/mol
by substitution:
q2 = 1 mol * 6.01 KJ/mol
= 6.01 KJ
3- the heat required to increase the water temperature from 0°C to 60 °C is:
q3 = m*C*ΔT
when m is the mass of water = 18 g
C is the specific heat capacity of water = 4.18 J/g-K
ΔT is the change of Temperature of water = 60°C - 0°C = 60°C
by substitution:
∴q3 = 18 g * 4.18 J/g-K * 60°C
= 4514 J = 4.514 KJ
∴the total change of enthalpy = q1+q2+q3
= 1.881 KJ +6.01 KJ + 4.514 KJ
= 12.405 KJ
