Answer:
First, we need to calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 85 °C - 35 °C
ΔT = 50 °C
Next, we can use the following formula to calculate the heat energy required:
Q = m·C·ΔT
where Q is the heat energy in calories, m is the mass of the sample in grams, C is the specific heat in cal/g °C, and ΔT is the change in temperature in °C.
Plugging in the given values, we get:
Q = 35.0 g · 0.108 cal/g °C · 50 °C
Q = 189.0 calories
Therefore, 189.0 calories are required to raise the temperature of the sample from 35 °C to 85 °C
Which of the following molecules is drawn in a conformation that has a proton and a leaving group anti-periplanar? H₂C, Br Ph. H CH3 Br H H₂C Br H₂C Ph H₂C CH3 H Ph H₂C, Br H Ph Save for Later CH3 CH3 CH3 CH3 CH3 Sul
The molecule that is drawn in a conformation that has a proton and a leaving group anti-periplanar is H₂C, Br.
The A, B, C, and D bond angles of a molecule are referred to as anti-periplanar, or antiperiplanar, in organic chemistry. The dihedral angles of the A–B and C–D bonds in this conformer are larger than +150° or less than 150°. In textbooks, the term "anti-periplanar" is frequently used to refer to a strictly anti-coplanar structure with a 180° AB CD dihedral angle. The anti-periplanar functional groups will be 180° apart from one another and in a staggered configuration in a Newman projection of the molecule.
Conformation is an essential factor in predicting reactivity in organic molecules. The anti-periplanar conformation of a molecule is one that occurs when two atoms in a molecule are in the same plane and are separated by 180 degrees. In this case, the proton and leaving group are placed in a perpendicular plane to the atoms directly in between them. This is the most stable conformer of the molecule. A significant factor in predicting reactivity in organic molecules is conformation. In this case, the molecule H₂C, Br is drawn in a conformation that has a proton and a leaving group anti-periplanar.
Therefore, the correct option is H₂C, Br.
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We know that Paz is trying to produce ammonia (NH3) from thin air. From looking at the experimental set-up, what are the reactants? a) NO2 and H20 b) N2 and H2 c) NO2 and H2 d) N2 and H20
To produce ammonia (NH₃) from thin air, the reactants required are N₂ and H₂. So the correct option is b).
Give a brief account on production of ammonia.Ammonia is one of the most abundantly produced inorganic chemicals. In 2016, there are a number of large ammonia plants around the world that produced a total of 144 million tons of nitrogen (equivalent to 175 million tons of ammonia). That number will rise to 235 million tonnes of ammonia in 2021. China produced 31.9% of its global production, followed by Russia at 8.7%, India at 7.5% and the United States at 7.1%. More than 80% of the ammonia produced is used as fertilizer for agricultural crops.
Today, most ammonia is produced on a large scale using the Haber process, with capacities of up to 3,300 tons per day. Gases N₂ and H₂ are reacted at a pressure of 200 bar. A typical modern ammonia production plant first converts natural gas, LPG, or petroleum gas into gaseous hydrogen. The process of producing hydrogen from hydrocarbons is known as steam reforming. Hydrogen then combines with nitrogen to produce ammonia by the Haber-Bosch process.
One way to produce green ammonia is to use hydrogen from the electrolysis of water and nitrogen separated from air. These are fed into the Haber Process (aka Haber-Bosch), all of which produce sustainable power.
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use the trendline equation in fig6.2 to determine the kelvin temperature at which the pressure equals .72 atm
When the pressure is 0.72 atm, the temperature in Kelvin is 156 K.
To determine the Kelvin temperature when the pressure is 0.72 atm, you will need to use the trendline equation given in Fig 6.2. First, find the equation of the trendline by using the graph's two points, (300 K, 1 atm) and (500 K, 2 atm).
The equation for the trendline is:
y = mx + b
Where y is pressure, x is the temperature in Kelvin, m is the slope, and b is the y-intercept. We can find the slope of the trendline by using the two points provided in the graph:
Slope (m) = (y2 - y1) / (x2 - x1)
Slope = (2 atm - 1 atm) / (500 K - 300 K)
Slope = 0.005 atm/K
The equation for the trendline can now be written: y = 0.005x + b. To find the y-intercept, b, we can use one of the two points: Solving for b:
1 atm = 0.005(300 K) + bb = 1.5 atm
Now we can use the equation for the trendline to find the temperature (x) at which the pressure (y) equals 0.72 atm:
0.72 atm = 0.005x + 1.5 atm
0.72 atm - 1.5 atm = 0.005x
-0.78 atm = 0.005xx
= -0.78 atm / 0.005x
= 156K
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Identify each of the following orbitals, and determine the n and quantum numbers. Explain your answers.
(a) one radial node the Number of radial nodes = n - l - 1
And number of angular nodes = l
n = 3 and l = 1
Orbital is 3p.
(b) It has zero angular node hence s-orbital and there is 1 radial node . 1 = n - 0 - 1 ; n = 2 and l = 0
The orbital is 2s.
(c) the shape of the orbital is that of dz². There is two angular nodes and there is no radial node.
n = 3 and l = 2
Hence the orbital is 3dz².
What is radial node?In atomic physics, a radial node is a point in space where the probability density of finding an electron in an atom is zero. It is a type of nodal plane that occurs in atomic orbitals, which are regions of space where electrons are most likely to be found.
Radial nodes occur in the radial distribution function of an atomic orbital, which describes the probability density of finding an electron at a given distance from the nucleus. The number of radial nodes in an atomic orbital is equal to n - l - 1, where n is the principal quantum number and l is the azimuthal quantum number.
Radial nodes represent regions of space where the radial wave function of the electron changes sign.
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