Answers
Answer:For the 1st box it starts with 250 and for the 2nd box it starts with CO(2).
Explanation:
Related Questions
Why does glucose and acentic acid have the same empirical formula
Answers
Answer:
Examples. Glucose (C6H12O6), ribose (C5H10O5), Acetic acid (C2H4O2), and formaldehyde (CH2O) all have different molecular formulas but the same empirical formula: CH2O.
Explanation:In chemistry, the empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound.
Can someone please help with these 2?
Answers
Equilibrium shifts to the right.
OPTION A
An infant acetaminophen suspension contains 80 mg/0.80 mL suspension. The recommended dose is 15 mg/kg body weight.
How many milliliters of this suspension should be given to an infant weighing 13 lb.
Answers
Answer:
0.8853 mL
Explanation:
First we convert 13 lb to kg, keeping in mind that 1 lb = 0.454 kg:
13 lb * [tex]\frac{0.454kg}{1lb}[/tex] = 5.902 kgThen we calculate how many mg of acetaminophen should be given, using the recommended dose and infant mass:
15 mg/kg * 5.902 kg = 88.53 mgFinally we calculate the required mL of suspension, using its concentration:
88.53 mg ÷ (80 mg/0.80 mL) = 0.8853 mLA 15.0 mL urine from a dehydrated patient has a density of 1.019g/mL. What is the mass of the sample, reported in mg?
Answers
Answer:
Mass of sample in mg = 15,285 mg
Explanation:
Given:
Volume of urine sample = 15 ml
Density of sample = 1.019 g/ml
FInd:
Mass of sample in mg
Computation:
Mass = density x volume
Mass of sample in mg = Volume of urine sample x Density of sample
Mass of sample in mg = 1.019 x 15
Mass of sample in mg = 15.285 gram
Mass of sample in mg = 15.285 x 1,000
Mass of sample in mg = 15,285 mg
1. Determine the volume of SO2 (at STP) formed from the reaction of 96.7 mol FeS2 and 55.0 L of O2 at 358 K and 1.20 atm.
4 FeS2(s) + 11O2(g) 2Fe2O3(s) + 8SO2(g)
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Answer:
40.0L of SO2 are produced
Explanation:
To solve this question we need to find the moles of O2 using PV = nRT in order to find the moles. Thus, we can find the limiting reactant and the moles (And volume) of SO2 produced as follows:
Moles O2:
n = PV/RT
n = 1.20atm*55.0L / 0.082atmL/molK*358K
n = 2.25 moles of O2.
Clearly, limiting reactant is O2.
The moles of SO2 produced are:
2.25 moles of O2 * (8mol SO2 / 11mol O2) = 1.6351 moles SO2
Volume SO2:
V = nRT/P
V = 1.6351 moles SO2*0.082atmL/molK*358K / 1.20atm
V = 40.0L of SO2 are produced
An elementary step is defined as a chemical collision in a reaction mechanism. A collection of different types of collisions makes up the reaction mechanism, so elementary steps provide a molecular view of the overall reaction.
a. True
b. False
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You are given a solution containing a pair of enantiomers (A and B). Careful measurements show that the solution contains 98% A and 2% B. What is the ee of this solution
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Answer:
ee = 96%
Explanation:
Enantiomeric excess, ee, is a way to express a mixture that is not enantiomerically pure. It is defined as 100 times the ratio between the differences of amounts of enantiomers and the total amunt. that is:
ee = |A-B|/ A+B * 100
ee = |98%-2%| / 98+2 * 100
ee = 96%How much energy does an X-ray with an 8 nm (8 x 10-9m) wavelength have?
A. 1.99 x 10-25 J
B. 3.33 x 1016 J
C. 2.48 x 10-17 j
D. 8.28 x 10-26 J
Answers
Answer:
it would be option C
Explanation:
Speed of light = 3×10^8m/s
Planck's constant = 6.626×10^-34 Js
Wavelength = 8 x 10^-9 m
Energy = [(3×10^8) * (6.626×10^-34)] / 8 x 10^-9
Energy = [19.878×10^(8-34)] / 8 x 10^-9
Energy = 2.48475 × 10^(-26+9)
Energy = 2.48×10^-17 J
This week's imide synthesis involves two reactions. In the first reaction (24A), a(n) ________ bond is formed between the two reactants. Hint: What type of functional group is formed
Answers
Answer:
C - N Bond formation.
Explanation:
Imide synthesis is a chemical reaction in organic chemistry which consists of two acyl groups which bond to nitrogen atom. The compound structure is related to acid anhydrides. Imides are monoacyl which are used as valuable intermediates in organic synthesis.
A diver exhales a bubble with volume of 250 mL at pressure of 2.4 atm and temperature of 15 C. How many gas particulate in this bubble?
Answers
Answer:
1.5x10²² particulates
Explanation:
Assuming ideal behaviour, we can solve this problem by using the PV=nRT formula, where:
P = 2.4 atmV = 250 mL ⇒ 250 / 1000 = 0.250 Ln = ?R = 0.082 atm·L·mol⁻¹·K⁻¹T = 15 °C ⇒ 15 + 273 = 288 KWe input the given data:
2.4 atm * 0.250 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 288 KAnd solve for n:
n = 0.025 molFinally we calculate how many particulates are there in 0.025 moles, using Avogadro's number:
0.025 mol * 6.023x10²³ particulates/mol = 1.5x10²² particulatesNaturally occurring gallium is a mixture of isotopes
that contains 90.11% of Ga-69 (atomic mass = 68.93
u) and 9.89% of Ga-71 (atomic mass 70.92 u).
What is the average atomic mass of naturally
occurring gallium?
A) 69.93 amu
C) 69.50 amu
B) 69.12 amu
D) 69.00 amu
