Answer:
This is the answer they're looking for:
Lysosomes and vacuoles both deal with waste materials. Lysosomes break down waste materials, and vacuoles store waste materials in the cell temporarily before the cell get rids of them.
Explanation:
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Answer:
The answer to the question is
The specific heat capacity of the alloy = 1.77 J/(g·°C)
Explanation:
To solve this, we list out the given variables thus
Mass of alloy = 45 g
Initial temperature of the alloy = 25 °C
Final temperature of the alloy = 37 °C
Heat absorbed by the alloy = 956 J
Thus we have
ΔH = m·c·(T₂ - T₁) where ΔH = heat absorbed by the alloy = 956 J, c = specific heat capacity of the alloy and T₁ = Initial temperature of the alloy = 25 °C , T₂ = Final temperature of the alloy = 37 °C and m = mass of the alloy = 45 g
∴ 956 J = 45 × C × (37 - 25) = 540 g·°C×c or
c = 956 J/(540 g·°C) = 1.77 J/(g·°C)
The specific heat capacity of the alloy is 1.77 J/(g·°C)
Cl2=3.17g/L
Ne=.901g/L
CO2=1.96g/l
therefore Cl2 is the densest gas under the given conditions.
Answer:
10.64
Explanation:
Let's consider the basic reaction of cyclohexamine, C₆H₁₁NH₂.
C₆H₁₁NH₂(aq) + H₂O(l) ⇄ C₆H₁₁NH₃⁺(aq) + OH⁻ pKb = 3.36
C₆H₁₁NH₃⁺ is its conjugate acid, since it donates H⁺ to form C₆H₁₁NH₂. C₆H₁₁NH₃⁺ acid reaction is as follows:
C₆H₁₁NH₃⁺(aq) + H₂O(l) ⇄ C₆H₁₁NH₂(aq) + H₃O⁺(aq) pKa
We can find the pKa of C₆H₁₁NH₃⁺ using the following expression.
pKa + pKb = 14.00
pKa = 14.00 - pKb = 14.00 - 3.36 = 10.64
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ANS:-Chloroplast
