Answer:
No, there are some organisms that only have one cell, although there are organ systems that are made up of many cells. Not all organisms have organ systems.
Explanation:
For example, the Trichoplax. It is an animal with no organs that is tiny and multicellular, and only feeds on microalgae. A Trichoplax are flat organisms about a millimetre in diameter with no internal structures and typically have two cellular layers. These organisms live in the oceans and seas all around the world but are not necessarily found in their natural habitat, more so in captivity such as a marine aquarium. Trichoplax are found to live in symbiosis with such things as bacteria and others in the oceans and seas.
Answer is: boiling point will be changed by 4°C.
Chemical dissociation of aluminium nitrate in water: Al(NO₃)₃ → Al³⁺(aq) + 3NO⁻(aq).
Change in boiling point: ΔT =i · Kb · b.
Kb - molal boiling point elevation constant of water is 0.512°C/m, this the same for both solution.
b - molality, moles of solute per kilogram of solvent., this is also same for both solution, because ther is same amount of substance.
i - Van't Hoff factor.
Van't Hoff factor for sugar solution is 1, because sugar do not dissociate on ions.
Van't Hoff factor for aluminium nitrate solution is approximately 4, because it dissociates on four ions (one aluminium cation and three nitrate anions). So ΔT is four times bigger.
Answer:
1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate
Explanation:
Step 1: Data given
Mass of sodium bicarbonate = 2.7 grams
Step 2: The balanced equation
HCl + NaHCO3 ⇔ NaCl + H2O + CO2
Step 3: Calculate moles NaHCO3
moles NaHCO3 =2.7 g / 84 g/mol= 0.032 moles
Step 4: Calculate moles HCl
For 1 mol NaHCO3 we need 1 mol HCl
For 0.032 moles NaHCO3 = 0.032 moles HCl
Step 5: Calculate mass HCl
Mass HCl = moles HCl * molar mass HCl
mass HCl = 0.032 * 36.46 g/mol= 1.17 grams
1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate
Answer:
V = 65.81 L
Explanation:
En este caso, debemos usar la expresión para los gases ideales, la cual es la siguiente:
PV = nRT (1)
Donde:
P: Presion (atm)
V: Volumen (L)
n: moles
R: constante de gases (0.082 L atm / mol K)
T: Temperatura (K)
De ahí, despejando el volumen tenemos:
V = nRT / P (2)
Sin embargo como estamos hablando de condiciones normales de temperatura y presión, significa que estamos trabajando a 0° C (o 273 K) y 1 atm de presión. Lo que debemos hacer primero, es calcular los moles que hay en 50 g de amoníaco, usando su masa molar de 17 g/mol:
n = 50 / 17 = 2.94 moles
Con estos moles, reemplazamos en la expresión (2) y calculamos el volumen:
V = 2.94 * 0.082 * 273 / 1
<h2>
V = 65.81 L</h2>
