Because that solid is frozen liquid, so when it gets heated up, it will melt. Like Ice.
Hope I Helped! :)
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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Answer:
The concentration of the solution is 1.364 molar.
Explanation:
Volume of perchloric acid = 29.1 mL
Mass of the solution = m
Density of the solution = 1.67 g/mL

Percentage of perchloric acid in 48.597 solution :70.5 %
Mass of perchloric acid in 48.597 solution :
= 
Moles of perchloric acid = 
In 29.1 mL of solution water is added and volume was changed to 250 mL.
So, volume of the final solution = 250 mL = 0.250 L (1 mL = 0.001 L)


The concentration of the solution is 1.364 molar.
B, millimeters because paper is really thin therefore it'd require small measurements for units.
Answer:
moles of glucose
<u>2.3166 moles of glucose</u>
<u></u>
Explanation:
The balance reaction for the formation of glucose is :

here , CO2 = carbon dioxide
H2O = water
C6H12O6 = glucose
O2 = Oxygen
According to this equation :
6 mole of CO2 = 6 mole of H2O = 1 mole of C6H12O6 = 6 mole of O2
We are asked to calculate the mole of Glucose from carbon dioxide.
So,
6 mole of CO2 produce = 1 mole of C6H12O6
1 mole of CO2 will produce =
moles of glucose
13.9 moles of CO2 will produce :

=2.3166 moles of glucose
Note : first , Always calculate for one mole (By dividing)
. After this , multiply the answer with the moles given.
Always write the substance whose amount is asked(glucose) to the right hand side