A contour line connects points of the same _____. A. latitude B. longitude C. elevation D. rock type

During nuclear fission, great amounts of energy are produced from____. A. particle accelerators B. tremendous amounts of mass C.

Phoenix [80]

During nuclear fission, great amounts of energy are produced from____.

Answer: Out of all the options presented above, the one that completes the statement and makes it true, is answer choice D) very small amounts of mass. Nuclear fission power does not produce carbon and also produces 12% of the world's power.

I hope it helps, Regards.

3 0

3 years ago

Read 2 more answers

A piece of lumber is 7.60cm long. What is it’s length in millimetres and in inches?

MrRissso [65]

Explanation:

Since 10mm is 1cm

Therefore 7.60cm is 760mm

Since 1 inch is 2.54cm

Therefore 7.60cm is 3.10 inches

4 0

1 year ago

What are two possible fates of stars larger than 8 sol

attashe74 [19]

Stars born larger than 8 solar masses usually retain enough mass to undergo core collapse, with the resulting shock wave producing a Type Ib supernova (spectra without Hydrogen or Silicon lines, with Helium lines), a Type Ic supernova (without Hydrogen or Helium or Silicon lines) or a Type II supernova

I hope it’s what you need

7 0

3 years ago

Sighting along the C2-C3 bond of 2-methylbutane, the least stable conformation (Newman projection) has a total energy strain of

natima [27]

Answer:

21 KJ/mol

Explanation:

For this question, we have to start with the linear structure of 2-methylbutane. With the linear structure, we can start to propose all the Newman projections keep it in mind that the point of view is between carbons 2 and 3 (see figure 1).

Additionally, we have several energy values for each interaction present in the Newman structures:

-) Methyl-methyl gauche: 3.8 KJ/mol

-) Methyl-H eclipse: 6.0 KJ/mol

-) Methyl-methyl eclipse: 11.0 KJ/mol

-) H-H eclipse: 4.0 KJ/mol

Now, we can calculate the energy for each molecule.

Molecule A

In this molecule, we have 2 Methyl-methyl gauche interactions only, so:

(3.8x2) = 7.6 KJ/mol

Molecule B

In this molecule, we have a Methyl-methyl eclipse interaction a Methyl-H eclipse interaction and an H-H eclipse interaction, so:

(11)+(6)+(4) = 21 KJ/mol

Molecule C

In this molecule, we have 1 Methyl-methyl gauche interaction only, so:

3.8 KJ/mol

Molecule D

In this molecule, we have three Methyl-H eclipse interaction, so:

(6*3) = 18 KJ/mol

Molecule E

In this molecule, we have 1 Methyl-methyl gauche interaction only, so:

3.8 KJ/mol

Molecule F

In this molecule, we have a Methyl-methyl eclipse interaction a Methyl-H eclipse interaction and an H-H eclipse interaction, so:

(11)+(6)+(4) = 21 KJ/mol

The structures with higher energies would be less stable. In this case, structures B and F with an energy value of 21 KJ/mol (see figure 2).

I hope it helps!

3 0

3 years ago

Calculate the maximum volume (in mL) of 0.143 M HCl that each of the following antacid formulations would be expected to neutral

Nuetrik [128]

Answer:

a. The maximum volume of 0.143 M HCl required is 154.4 mL.

b. The maximum volume of 0.143 M HCl required is 135.7 mL.

Explanation:

a.

$Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O$

Mass of aluminum hydroxide = 350 mg = 0.350 g ( 1mg = 0.001 g)

Moles of aluminum hydroxide = $\frac{0.350 g}{78 g/mol}=0.004487 mol$

According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :

$\frac{3}{1}\times 0.004487 mol=0.01346 mol$ of HCl.

$Mg(OH)_2+2HCl\rightarrow MgCL_2+2H_2O$

Mass of magnesium hydroxide = 250 mg = 0.250 g ( 1mg = 0.001 g)

Moles of magnesium hydroxide = $\frac{0.250 g}{58 g/mol}=0.004310 mol$

According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310 mole of magnesium hydroxide will be neutralize by :

$\frac{2}{1}\times 0.004310 mol=0.008621 mol$ of HCl.

Total moles of HCl required to neutralize both :

0.01346 mol + 0.008621 mol = 0.02208 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

$Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}$

$V=\frac{0.02208 mol}{0.143 M}=0.1544 L$

1 L = 1000 mL

0.1544 L = 154.4 mL

The maximum volume of 0.143 M HCl required is 154.4 mL.

b.

$CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2$

Mass of calcium carbonate = 970mg = 0.970 g ( 1mg = 0.001 g)

Moles of calcium carbonate = $\frac{0.970 g}{100 g/mol}=0.00970 mol$

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

$\frac{2}{1}\times 0.00970 mol=0.0194 mol$ of HCl.

Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

$Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}$

$V=\frac{0.0194 mol}{0.143 M}=0.1357 L$

1 L = 1000 mL

0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.

4 0

3 years ago