Volume of Hydrogen V1 = 351mL
Temperature T1 = 20 = 20 + 273 = 293 K
Temperature T2 = 38 = 38 + 273 = 311 K
We have V1 x T2 = V2 x T1
So V2 = (V1 x T2) / T1 = (351 x 311) / 293 = 372.56
Volume at 38 C = 373 ml
The individual can consume less than 184.6 g of the snack mix and still be within the FDA limit of salt consumption.
<h3>What is the mass of snack that can be consumed within the limit of sodium intake?</h3>
The mass of the snack mix that the individual can consume and still be within the FDA limit is calculated as follows:
U.S. Food and Drug Administration (FDA) recommends of sodium intake = less than 2.40 g of sodium per day.
Amount of salt in 100 g of snack mix = 1.30 g
Mass of snack that will contain 2.40 g of sodium = 2.40 * 100g/1.30 = 184.6 g of snack mix
Therefore, the individual can consume less than 184.6 of the snack mix and still be within the FDA limit of salt consumption.
In conclusion, the FDA recommends that an individual take in less than 2.40 g of sodium per day from their diet.
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Answer:
The answer to your question is Q = 355.64 J
Explanation:
Data
Heat = Q = ?
Temperature 1 = T1 = 20°C
Temperature 2 = T2 = 37°C
mass = m = 5 g
Specific heat = Cp = 4.184 J/g°C
Formula
Q = mCp(T2 - T1)
-Substitution
Q = (5)(4.184)(37 - 20)
-Simplification
Q = (5)(4.184)(17)
-Result
Q = 355.64 J
Answer:
1.26 M
Explanation:
The ion nitrate is NO₃⁻ and the Barium is from group 2 so it forms the ion Ba²⁺, so the barium nitrate has the formula: Ba(NO₃)₂. The molar masses are: Ba: 137 g/mol, N = 14 g/mol, O = 16 g/mol, so the molar mass of barium nitrate is:
137 + 2x(14 + 3x16) = 199 g/mol
The number of moles is the mass divided by the molar mass, so:
n = 25.1/199 = 0.126 mol of Ba(NO₃)₂
In 1 mol of the salt, there are 2 moles of NO₃⁻, so the number of moles of nitrate is 0.252 mol. Nitrates formed with ammonium (that can react when the solid dissolves) and with elements from group 1 and 2 are completely soluble in water. So, the moles of nitrate will remain 0.252 mol.
The molarity is the number of moles divided by the volume (0.2 L):
[NO₃⁻]= 0.252/0.2 = 1.26 M
