Silicon is lustrous and is a conductor of heat and electricity. Which class of elements could this element belong to?

What does the Periodic Table show?

Mamont248 [21]

Atomic number
atomic mass
group|family
periods
and element symbols

8 0

3 years ago

If the volume and Kelvin temperature are reduced by one-half what will happen to the pressure???

jolli1 [7]

Answer:

the pressure will decrease

Explanation:

6 0

3 years ago

All living things are made up of molecules that have carbon. O True O False

Paha777 [63]

Answer:

True

Explanation:

Because Carbon is the primary component of macromolecules, including proteins, lipids, nucleic acids, and carbohydrates.

6 0

2 years ago

Read 2 more answers

Oxana [17]

Answer:

1422mg of acetaminophen

Explanation:

The maximum dose of acetaminophen is 15.0 mg of acetaminophen per kg of person.

To know the maximum single dosage of the person we need to convert the 209lb to kg (Using 1kg = 2.2046lb):

209lb * (1kg / 2.2046lb) = 94.8

The person weighs 94.8kg and the maximum single dosage for the person is:

94.8kg * (15.0mg acetaminophen / kg) =

1422mg of acetaminophen

4 0

2 years ago

Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e

igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

Explanation:

1) Moles of NaCl , $n_1=1 mol$

Mass of water = m= 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute(NaCl)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p=17.19 Torr$

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose , $n_1=1mol$

Mass of water = m = 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p'$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute ( glucose)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p'=17.19 Torr$

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

3 0

3 years ago