The change in temperature (ΔT) : 56.14 ° C
<h3>Further explanation</h3>
Given
Cereal energy = 235,000 J
mass of water = 1000 g
Required
the change in temperature (ΔT)
Solution
Heat can be formulated :
Q = m . c . ΔT
c = specific heat for water = 4.186 J / gram ° C
235000 = 1000 . 4.186 . ΔT
Answer: This is from a wiki i found. Approximately one third of a cell’s proteins are destined to function outside the cell’s boundaries or while embedded within cellular membranes. Ensuring these proteins reach their diverse final destinations with temporal and spatial accuracy is essential for cellular physiology. In eukaryotes, a set of interconnected organelles form the secretory pathway, which encompasses the terrain that these proteins must navigate on their journey from their site of synthesis on the ribosome to their final destinations. Traffic of proteins within the secretory pathway is directed by cargo-bearing vesicles that transport proteins from one compartment to another. Key steps in vesicle-mediated trafficking include recruitment of specific cargo proteins, which must collect locally where a vesicle forms, and release of an appropriate cargo-containing vessel from the donor organelle (Figure 1). The newly formed vesicle can passively diffuse across the cytoplasm, or can catch a ride on the cytoskeleton to travel directionally. Once the vesicle arrives at its precise destination, the membrane of the carrier merges with the destination membrane to deliver its cargo. Have a nice day.
Explanation: Plz make brainliest
Stoichiomety:
1 moles of C + 1 mol of O2 = 1 mol of CO2
multiply each # of moles times the atomic molar mass of the compund to find the relation is weights
Atomic or molar weights:
C: 12 g/mol
O2: 2 * 16 g/mol = 32 g/mol
CO2 = 12 g/mol + 2* 16 g/mol = 44 g/mol
Stoichiometry:
12 g of C react with 32 g of O2 to produce 44 g of CO2
Then 18 g of C will react with: 18 * 32/ 12 g of Oxygen = 48 g of Oxygen
And the result will be 12 g of C + 48 g of O2 = 60 g of CO2.
You cannot obtain 72 g of CO2 from 18 g of C.
May be they just pretended that you use the law of consrvation of mass and say that you need 72 g - 18g = 54 g. But it violates the proportion of C and O2 in the CO2 and is not possible.
H2SO.Mgslfurmobile phase in this experiment
Answer:
Mass = 0.32 g
Explanation:
Given data:
Mass of CH₄ = ?
Volume of CH₄ = 500 mL (500 mL× 1L/1000 mL= 0.5 L)
Temperature = 273 K
Pressure = 1 atm
Solution:
Volume of CH₄:
500 mL (500 mL× 1L/1000 mL= 0.5 L)
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
By putting values,
1 atm× 0.5 L = n×0.0821 atm.L/ mol.K × 273 K
0.5 atm.L = n×22.4 atm.L/ mol
n = 0.5 atm.L / 22.4 atm.L/ mol
n = 0.02 mol
Mass in gram:
Mass = number of moles × molar mass
Mass = 0.02 mol × 16 g/mol
Mass = 0.32 g
