Answer:
The answer to your question is T2 = 319.4°K
Explanation:
Data
Volume 1 = V1 = 852 ml
Temperature 1 = T1 = 288°K
Volume 2 = V2 = 945 ml
Temperature 2 = T2 = ?
Process
To solve this problem, use Charles' law.
V1/T1 = V2/T2
-Solve for T2
T2 = V2T1/V1
-Substitution
T2 = (945 x 288) / 852
-Simplification
T2 = 272160 / 852
-Result
T2 = 319.4°K
Well protists quite very as organisms i.e it is hard to say if an organism is a protist. Some main features of protists are that they have
1. A membrane enclosed nucleus
2.They are mainly single celled
Also protista are found mainly anywhere water is present for example damp soil, pools and lakes.
Hope this helps :).
A liquid with high viscosity does not flow easily and is not effective in wetting a surface.
When a metal is subjected to corrosive elements including salt, moisture, and high temperatures, a reaction called corrosion takes place inside the metal. Some foods contain metallic compounds that can corrode a material. The majority of corrosion is simply surface dis-colouration, which polishing agents may quickly remove.
Increasing viscosity and constant intermolecular water bonding together result in surface tension. Any liquid that was more viscous than water possessed a surface tension that was equal to or lower than that of water. Viscosity with surface tension decrease when temperature rises.
Therefore, a liquid with high viscosity does not flow easily and is not effective in wetting a surface.
To know more about viscosity
brainly.com/question/2193315
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The net ionic equation is
Cu(s) + 4H⁺(aq) + 4NO₃⁻(aq) ⟶ Cu²⁺(aq) + 2NO₃⁻(aq) + 2NO₂(g) + 2H₂O(ℓ)
<em>Molecular equation
:</em>
Cu(s) + 4HNO₃(aq) ⟶ Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(ℓ)
<em>Ionic equation:
</em>
Cu(s) + 4H⁺(aq) + 4NO₃⁻(aq) ⟶ Cu²⁺(aq) + 2NO₃⁻(aq) + 2NO₂(g) + 2H₂O(ℓ)
<em>Net ionic equation
</em>
Cu(s) + 4H⁺(aq) + 4NO₃⁻(aq) ⟶ Cu²⁺(aq) + 2NO₃⁻(aq) + 2NO₂(g) + 2H₂O(ℓ)
<em>Note</em>: The net ionic equation is <em>the same as </em>the ionic equation because there are <em>no common ions</em> to cancel on opposite sides of the arrow.
