Answer:
48.32 g of anhydrous MnSO4.
Explanation:
Equation of dehydration reaction:
MnSO4 •4H2O --> MnSO4 + 4H2O
Molar mass = 55 + 32 + (4*16) + 4((1*2) + 16)
= 223 g/mol
Mass of MnSO4 • 4H2O = 71.6 g
Number of moles = mass/molar mass
= 71.6/223
= 0.32 mol.
By stoichiometry, since 1 mole of MnSO4 •4H2O is dehydrated to give 1 mole of anhydrous MnSO4
Number of moles of MnSO4 = 0.32 mol.
Molar mass = 55 + 32 + (4*16)
= 151 g/mol.
Mass = 151 * 0.32
= 48.32 g of anhydrous MnSO4.
Answer:
The water lost is 36% of the total mass of the hydrate
Explanation:
<u>Step 1:</u> Data given
Molar mass of CuSO4*5H2O = 250 g/mol
Molar mass of CuSO4 = 160 g/mol
<u>Step 2:</u> Calculate mass of water lost
Mass of water lost = 250 - 160 = 90 grams
<u>Step 3:</u> Calculate % water
% water = (mass water / total mass of hydrate)*100 %
% water = (90 grams / 250 grams )*100% = 36 %
We can control this by the following equation
The hydrate has 5 moles of H2O
5*18. = 90 grams
(90/250)*100% = 36%
(160/250)*100% = 64 %
The water lost is 36% of the total mass of the hydrate
The answer is A to B because the distance is rising rapidly as seen by the steep slope segment A to B had
Answer:
oof idk I would suggest looking at an example and try going off that
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
