Response:
2200 mg of antibiotic
Explanation:
The prescribed antibiotic dosage is 40 mg/kg of body weight.
For a patient weighing 55 kg, we calculate the dose of antibiotic as follows:
If we analyze 40/1000000, we can determine antibiotic allocation in kg per kg of weight
= 0.00004 kg of antibiotic for each kilogram
0.00004 multiplied by 55 (to find out the required amount for a 55 kg individual)
= 0.0022 kg
This 0.0022 figure converts to milligrams as follows
0.0022*10^6
= 2200 mg of antibiotic is indicated for a patient weighing 55 kg.
Utilize the ideal gas law:
n = PV / RT
P = 100kPa = 100 x 1000 x (9.8 x 10^{-6}) = 0.98 atm
Convert kPa to atm, where 1 Pa = 9.8 x 10^{-6} atm.
T = 293 K
V = 6.8 L
R = 1/12
Substituting all values leads to:
n = 0.272
Answer:
C) 1.15 × 10⁻⁷ mm
Explanation:
Step 1: Provided information
Average separation between oxygen and nitrogen atoms: 115 pm
Step 2: Change the distance to meters (SI standard unit)
Using the conversion 1 m = 10¹² pm.
115 pm × (1 m/10¹² pm) = 1.15 × 10⁻¹⁰ m
Step 3: Transform the distance to millimeters
Employing the conversion 1 m = 10³ mm.
1.15 × 10⁻¹⁰ m × (10³ mm/1 m) = 1.15 × 10⁻⁷ mm
Answer:
Explanation:
In KCl, the two elements that combine to create KCl are potassium (K) and chlorine (Cl).
Potassium, as a Group 1 element, possesses one valence electron in its outermost shell which it readily donates during bonding. Every element aims to achieve a stable electron configuration, typically with 2 or 8 electrons in its outer shell. Potassium is characterized by its lower electronegativity and higher ionization energy, making it more likely to donate its electron than to accept one. On the other hand, chlorine belongs to Group 17 and has 7 electrons in its outer shell, requiring just one additional electron to complete its octet. Chlorine’s higher electronegativity and lower ionization energy facilitate its tendency to accept an electron rather than donate it.
The bond between potassium and chlorine that results in KCl is termed an electrovalent bond.
Reaction equation:
K + Cl → KCl
Answer:
The temperature difference is 293.15 Kelvin.
Explanation:
The provided information:
The temperature difference between the two matters is 20°C
We need to determine this difference in Kelvin =?
To solve this;
Using the formula:
0°C +273.15
Now substituting the values in place of 0.
20°C + 273.15 = 293.15 K
Hence, the temperature differential between the two samples is 293.15 K.
