Cobalt is a metal thus the atoms are bonded together in giant atomic structures and metallic bonds. The atoms attract each other with very strong atomic forces this the resulting structure is giant atomic structures. This provides for the extra strong structure of cobalt.
The type of the bond is present Na₃PO₄ is the ionic bond. the Na₃PO₄ is the ionic compound. yes the Na₃PO₄ is the polyatomic ion.
The Na₃PO₄ is Na⁺ and PO₄³⁻. the phosphorus is the non metal and the oxygen atom is the non metal. the non meta and non meta form the covalent or molecular bond. the bond between the PO₄³⁻ bond is the covalent bond but the overall present in the Na₃PO₄ is the ionic bond . the bons in between the Na⁺ and PO₄³⁻ is the the ionic bond. the PO₄³⁻ id the polyatomic ion .
The bond between the positively charged ion and the negatively charged ion are called as the ionic bond and the compound form is the ionic compound.
To learn more about ionic bond here
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Each enzyme's active site is suitable for one specific type of substrate – just like a lock that has the right shape for only one specific key. Changing the shape of the active site of an enzyme will cause its reaction to slow down until the shape has changed so much that the substrate no longer fits.
I don't know what the problem is, but here are some rues to help you out:
- All non-zero figures are significant
- When a zero falls between non-zero digits, that zero is significant.
- When a zero falls after a decimal point, that zero is significant.
- When multiplying and dividing significant figures, the answer is limited to the number of sig figs equal to the least number of sig figs in the problem.
- When adding and subtracting, the answer is limited to the number of decimal places in the number with the least number of decimal places.
Answer:
31395 J
Explanation:
Given data:
mass of water = 150 g
Initial temperature = 25 °C
Final temperature = 75 °C
Energy absorbed = ?
Solution:
Formula:
q = m . c . ΔT
we know that specific heat of water is 4.186 J/g.°C
ΔT = final temperature - initial temperature
ΔT = 75 °C - 25 °C
ΔT = 50 °C
now we will put the values in formula
q = m . c . ΔT
q = 150 g × 4.186 J/g.°C × 50 °C
q = 31395 J
so, 150 g of water need to absorb 31395 J of energy to raise the temperature from 25°C to 75 °C .
