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1 year ago

Calculate the mass (in grams) of 250mL of ether at 25 oC. The density of

Chemistry

1 answer:

leonid [27]1 year ago

6 0

Volume=250mL
Density=0.71g/ml

$\boxed{\sf Density=\dfrac{Mass}{Volume}}$

$\\ \sf{:}\implies Mass=Density(Volume)$

$\\ \sf{:}\implies Mass=0.71(250)$

$\\ \sf{:}\implies Mass=177.5g$

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When a known quantity of compound, at a known concentration, is added to a known volume of another compound to determine the con

Vladimir [108]

Answer:

A titration

Explanation:

A common example of a titration is when we have an acid of unknown concentration, so we add a known volume of a base of known concentration. This process lets us determine the concentration of the acid.

By definition, a titration is a quantitative analysis, as we determine how much of an analyte is there in a sample. However, there are quantitative analyzes which are not titrations. This is why the most appropiate answer is a titration.

5 0

2 years ago

Write the electron configurations for the following ions:

Ket [755]

Answer:

Co²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷

Sn²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰

Zr⁴⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶

Ag⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d¹⁰

S²⁻ : 1s² 2s² 2p⁶ 3s² 3p⁶

Explanation:

Cobalt (Co): atomic number 27

The electronic configuration of Co in ground state:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁷

The electronic configuration of Co in +2 oxidation state (Co²⁺) :

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷

Tin (Sn): atomic number 50

The electronic configuration of Sn in ground state:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p²

The electronic configuration of Sn in +2 oxidation state (Sn²⁺) :

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰

Zirconium (Zr): atomic number 40

The electronic configuration of Zr in ground state:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d²

The electronic configuration of Zr in +4 oxidation state (Zr⁴⁺) :

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶

Silver (Ag): atomic number 47

The electronic configuration of Ag in ground state:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹ 4d¹⁰

The electronic configuration of Ag in +1 oxidation state (Ag⁺) :

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d¹⁰

Sulphur (S): atomic number 16

The electronic configuration of S in ground state:

1s² 2s² 2p⁶ 3s² 3p⁴

The electronic configuration of S in -2 oxidation state (S²⁻) :

1s² 2s² 2p⁶ 3s² 3p⁶

8 0

2 years ago

Read 2 more answers

A gas sample occupies 3.50 liters of volume at 20.°c. what volume will this gas occupy at 100.°c (reported to three significant

VLD [36.1K]

Explanation:

According to Charle's law, at constant pressure the volume of an ideal gas is directly proportional to the temperature.

That is, $Volume \propto Temperature$

Hence, it is given that $V_{1}$ is 3.50 liters, $T_{1}$ is 20 degree celsius, and $T_{2}$ is 100 degree celsius.

Therefore, calculate $V_{2}$ as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$

$\frac{3.50 liter}{20^{o}C} = \frac{V_{2}}{100^{o}C}$

$V_{2}$ = 17.5 liter

Thus, we can conclude that volume of gas required at 100 degree celsius is 17.5 liter.

6 0

2 years ago

Read 2 more answers

A beach has a supply of sand grains composed of calcite, ferromagnesian silicate minerals, and non-ferromagnesian silicate miner

bezimeni [28]

Ferromagnesian silicate minerals (i looked it up)

4 0

2 years ago

A bottle of concentrated aqueous sulfuric acid, labeled 98.0 wt% h2so4, has a concentration of 18.0 m. (a) how many milliliters

nadya68 [22]

n this order, Ď=1.8gmL, cm=0.5, and mole fraction = 0.9 First, let's start with wt%, which is the symbol for weight percent. 98wt% means that for every 100g of solution, 98g represent sulphuric acid, H2SO4. We know that 1dm3=1L, so H2SO4's molarity is C=nV=18.0moles1.0L=18M In order to determine sulphuric acid solution's density, we need to find its mass; H2SO4's molar mass is 98.0gmol, so 18.0moles1Lâ‹…98.0g1mole=1764g1L Since we've determined that we have 1764g of H2SO4 in 1L, we'll use the wt% to determine the mass of the solution 98.0wt%=98g.H2SO4100.0g.solution=1764gmasssolutionâ†’ masssolution=1764gâ‹…100.0g98g=1800g Therefore, 1L of 98wt% H2SO4 solution will have a density of Ď=mV=1800g1.0â‹…103mL=1.8gmL H2SO4's molality, which is defined as the number of moles of solute divided by the mass in kg of the solvent; assuming the solvent is water, this will turn out to be cm=nH2SO4masssolvent=18moles(1800â’1764)â‹…10â’3kg=0.5m Since mole fraction is defined as the number of moles of one substance divided by the total number of moles in the solution, and knowing the water's molar mass is 18gmol, we could determine that 100g.solutionâ‹…98g100gâ‹…1mole98g=1 mole H2SO4 100g.solutionâ‹…(100â’98)g100gâ‹…1mole18g=0.11 moles H2O So, H2SO4's mole fraction is molefractionH2SO4=11+0.11=0.9

5 0

2 years ago