For #2: Odor, Temperature, and color
The dermal tissue is basically the outer part of the cell that protects the plant cell so in that case it would be the carbohydrate or cell wall. If you would put the choices it could have been a bit better :)
Answer:
Weigh the empty crucible, and then weigh into it between 2 g and 3 g of hydrated copper(II) sulphate. Record all weighings accurate to the nearest 0.01 g.
Support the crucible securely in the pipe-clay triangle on the tripod over the Bunsen burner.
Heat the crucible and contents, gently at first, over a medium Bunsen flame, so that the water of crystallisation is driven off steadily. The blue colour of the hydrated compound should gradually fade to the greyish-white of anhydrous copper(II) sulfate. Avoid over-heating, which may cause further decomposition, and stop heating immediately if the colour starts to blacken. If over-heated, toxic or corrosive fumes may be evolved. A total heating time of about 10 minutes should be enough.
Allow the crucible and contents to cool. The tongs may be used to move the hot crucible from the hot pipe-clay triangle onto the heat resistant mat where it should cool more rapidly.
Re-weigh the crucible and contents once cold.
Calculation:
Calculate the molar masses of H2O and CuSO4 (Relative atomic masses: H=1, O=16, S=32, Cu=64)
Calculate the mass of water driven off, and the mass of anhydrous copper(II) sulfate formed in your experiment
Calculate the number of moles of anhydrous copper(II) sulfate formed
Calculate the number of moles of water driven off
Calculate how many moles of water would have been driven off if 1 mole of anhydrous copper(II) sulfate had been formed
Write down the formula for hydrated copper(II) sulfate.
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Explanation:
Answer:
Final molarity of iodide ion C(I-) = 0.0143M
Explanation:
n = (m(FeI(2)))/(M(FeI(2))
Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol
So n = 0.981/309.85 = 0.0031 mol
V(solution) = 150mL = 0.15L
C(AgNO3) = 35mM = 0.035M = 0.035m/L
n(AgNO3) = C(AgNO3) x V(solution)
= 0.035 x 0.15 = 0.00525 mol
(AgNO3) + FeI(3) = AgI(3) + FeNO3
So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol
C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M
Respuesta:
2400 mL
Explicación:
Paso 1: Información dada
- Volumen de solución: 3 L (3000 mL)
- Concentración de naranja: 20 % v/v
Paso 2: Calcular el volumen de naranja
La concentración de naranja es de 20 % v/v, es decir, cada 100 mL de solución hay 20 mL de naranja.
3000 mL Sol × 20 mL Naranja/100 mL Solución = 600 mL Naranja
Paso 3: Calcular el volumn de agua
El volumen de soluciónes igual a la suma de los volúmenes de naranja y agua.
VSolución = VNaranja + VAgua
VAgua = VSolución - VNaranja
VAgua = 3000 mL - 600 mL = 2400 mL
