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3 years ago

1.50μm2. Convert this to square meters.

1 answer:

6 0

Based on the standards of units conversion, to convert from micrometer to meter, we multiply by 10^-6.
Since we there is a square (10^2) to consider, then to convert from micrometer squared to meter squared, we will multiply by (10^-6)^2 as follows:
1.5 μm2 = 1.5 x (10^-6)^2 = 1.5 x 10^-12 meter sqaures

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Use bond energies from Table 10.3 in the textbook to estimate the enthalpy change (ΔH) for the following reaction. C2H2(g)+H2(g)

Digiron [165]

Answer: $=176.6kJmol^{-1}$

Explanation:Bond energy of H-H is 436.4 kJ/mole

Bond energy of C-H is 414 kJ/mol

Bond energy of C=C is 620 kJ/mol

Bond energy of C≡C is 835 kJ/mol

$\Delta H= {\text {sum of bond energies of reactants}}- {\text {sum of bond energies of products}}$

$\Delta H$ = {1B.E(C≡C)+2B.E(C-H) +1B.E(H-H)} - {1B.E(C=C)+4B.E(C-H)}

$\Delta H= {1B.E(835kJmole^{-1})+2B.E(414kJmole^{-1}) +1B.E(436.4kJmole^{-1})} - {1B.E(620kJmole^{-1})+4B.E(414kjmole^{-1})}$

$=176.6kJmol^{-1}$

7 0

3 years ago

calculate pressure exerted by 1.255 mol of CI2 in a volume of 5.005 L at a temperature 273.5 k using ideal gas equation

balu736 [363]

Answer:

The pressure is 5.62 atm.

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

P= ?
V= 5.005 L
n= 1.255 mol
R= 0.082 $\frac{atm*L}{mol*K}$
T= 273.5 K

Replacing:

P* 5.005 L= 1.255 mol* 0.082 $\frac{atm*L}{mol*K}$ *273.5 K

Solving:

$P=\frac{1.255 mol* 0.082 \frac{atm*L}{mol*K} *273.5 K}{5.005 L}$

P= 5.62 atm

The pressure is 5.62 atm.

8 0

2 years ago

Calcium and bromine have formed a bond. Leading up to this, calcium gave up electrons. It was a(n)

scZoUnD [109]

Leading up to this, calcium gave up 2 valence electrons and thus was denoted as a cation. These 2 electrons were transferred to bromine, which received an overall negative charge because of the addition of 2 valence electrons in its valence shell, and thus formed a negatively charged ion, an anion.

Both formed an ionic bond, due to the electrostatic charge of attraction between the 2 oppositely charged ions. If many ions of Ca and Br are present and numerous ionic bonds have formed it will undergo an arrangement which is that of an ionic lattice, type of structure.

6 0

3 years ago

During strenuous exercise, the NADH formed in the glyceraldehyde 3-phosphate dehydrogenase reaction in skeletal muscle must be r

vekshin1

Answer:

The answer is pyruvate → lactate

Explanation:

In the reaction of glycolysis, glucose breaks down to form pyruvate yielding ATP and NADH.

Under or during strenuous exercise, which is an anaerobic condition, lactate is formed by the reoxidization of NADH and the conversion of pyruvate to lactate.

6 0

2 years ago

3mL of cyclohexanol (density = 0.9624 g/mL, Molecular weight = 100.158 g/mol) reacts with excess sulfuric acid to produce cycloh

TiliK225 [7]

Answer:

$n_{C_6H_{10}}=0.03molC_6H_{10}$

Explanation:

Hello,

In this case the undergoing chemical reaction is shown on the attached picture whereas cyclohexanol is converted into cyclohexene and water by the dehydrating effect of the sulfuric acid. Thus, for the starting 3 mL of cyclohexanol, the following stoichiometric proportional factor is applied in order to find the theoretical yield of cyclohexene in moles:

$n_{C_6H_{10}}=3mLC_6H_{12}O*\frac{0.9624gC_6H_{12}O}{1mLC_6H_{12}O}*\frac{1molC_6H_{12}O}{100.158gC_6H_{12}O}*\frac{1molC_6H_{10}}{1molC_6H_{12}O} \\n_{C_6H_{10}}=0.03molC_6H_{10}$

Besides, the mass could be computed as well by using the molar mass of cyclohexene:

$m_{C_6H_{10}}=0.03molC_6H_{10}*\frac{82.143gC_6H_{10}}{1molC_6H_{10}} \\\\m_{C_6H_{10}}=2.4gC_6H_{10}$

Even thought, the volume could be also computed by using its density:

$V_{C_6H_{10}}=2.4gC_6H_{10}*\frac{1mLC_6H_{10}}{0.811gC_6H_{10}} \\V_{C_6H_{10}}=3mLC_6H_{10}$

Best regards.

4 0

2 years ago

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