What element is necessary for a healthy eye

If the density of a liquid 102g/cm^3, how many milligrams will be in 63 mL of that liquid?

Ede4ka [16]

Answer:

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

mass = Density × volume

From the question

63 mL = 63 cm³

We have

mass = 102 × 63 = 6426

But 1 g = 1000 mg

6426 g = 6426000 mg

We have the final answer as

Hope this helps you

6 0

3 years ago

Determine whether each of the properties described applies to volumetric or graduated glassware. 1. Used for applications in whi

kogti [31]

Answer:

1) volumetric

2) graduated

3) volumetric

Explanation:

A volumetric glassware is a glassware that is marked at a particular point. A typical example of a volumetric glassware is the volumetric flask. A volumetric glassware is capable of measuring only a specific volume of a liquid.

On the other hand, graduated glassware can measure a range of volumes of liquid. However, a volumetric glassware is still required where a high degree of accuracy is important.

5 0

3 years ago

How many carbon atoms would be in the compound named chlorobenzene

kondaur [170]

chlorobenzene
Carbon - 6
Hydrogen - 5
Chlorine - 1

that 1 chlorine replaces one of the hydrogens
thats why hydrogen number decreases by number of Cl atoms (that are substituting those H atoms)

7 0

3 years ago

1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.

Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice ) = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

$\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}$

$\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}$

$\Delta \ S = {(1*37.7)In \dfrac{263}{273}$

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0

3 years ago

HELP FOR FINAL

Dahasolnce [82]

Volume of CO₂ obtained : 13 L

<h3>Further explanation</h3>

Reaction

C₂H₅OH + 3 O₂ ⇒ 2 CO₂ + 3 H₂0 + 8842 Joules

moles of ethanol=0.28

From equation, mol ratio ethanol : CO₂ = 1 : 2, so mol CO₂ :

$\tt \dfrac{2}{1}\times 0.28=0.56$

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

Then volume of CO₂ :

$\tt V~CO_2=mol\times 22.4=0.56\times 22.4=12.544\approx 13~L$

5 0

2 years ago