What is the molar mass of Agl2

15.1 L N2 at 25 °C and 125 kPa and 44.3 L O2 at 25 °C and 125 kPa were transferred to a tank with a volume of 6.25 L. What is th

Y_Kistochka [10]

Answer:

The total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.

Explanation:

For N2,

Pressure(P₁)=125 kPa

Volume(V₁)=15·1 L

Temperature (T₁)=25°C=25+273 K=298 K

Similarly, for Oxygen,

Pressure(P₂)= 125 kPa

Volume(V₂)= 44.3 L

Temperature(T₂)=25°C= 298 K

Then, for the mixture,

Volumeof the mixture( V)= 6.25 L

Pressure(P)=?

Temperature (T)= 51°C = 51+273 K=324 K

Then, By Combined gas laws,

$\frac{P_{1} V_{1} }{T_{1} } +\frac{P_{2} V_{2} }{T_{2} } =\frac{PV}{T}$

or, $\frac{15.1*125}{298} +\frac{44.3*125}{298} =\frac{P*6.25}{324}$

or, $6.34+18.58=\frac{P*6.25}{324}$

or, $P=\frac{24.92*324}{6.25}$

∴P=1291.85 kPa

So the total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.

3 0

3 years ago

State whether homgenous mixture or heterogenoues mixture soda water

Citrus2011 [14]

Answer: homogeneous

Explanation: The taste of soda water is constant till the end. The carbon dioxide is dissolved homogeneously in soda water.

4 0

3 years ago

Which is a saturated solution?

ser-zykov [4K]

Answer:

C）52g KCl in 100g water at 80°C

Explanation:

A saturated solution is one that contains as much solute as it can dissolve in the presence of excess solute at that particular temperature.

A solutibility curve is a graph that shows the variability with temperature of the solubility of a solute in a given solvent. A solutibility curve can provide information of whether a solution formed frommthe solute and solvent are saturated or not at a given temperature.

From the solubility curve in the attachment below:

A) A saturated solution of NH₄Cl will contain about 52 g solute per 100 g sat 50 °C. Thus, a solution of 40 g NH₄Cl in 100 g water at 50 °C is an unsaturated solution.

B) A saturated solution of SO₂ at 10°C will contain about 70 g of solute in 100 g of water. Thus a solution of 2g SO₂ in 100g water at 10°C is an unsaturated solution.

C) A saturated solution of KCl at 80 °C will contain about 52 g of solute in 100 g of water. Thus, a solution of 52g KCl in 100g water at 80°C is a saturated solution.

D) A saturated solution of Kl at 20 °C will contain about 145 g of solute in 100 g of water. Thus, a solution of 120g KI in 100g water at 20°C is an unsaturated solution.

7 0

3 years ago

What is the mass of 4.99×1021 platinum atoms?

charle [14.2K]

Answer:

$\boxed {\boxed {\sf 1.62 \ g \ Pt}}$

Explanation:

We are asked to find the mass of a number of platinum (Pt) atoms.

<h3>1. Convert Atoms to Moles </h3>

First, we convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are platinum atoms.

We will convert using dimensional analysis so we set up a ratio using the information we know (6.022 × 10²³ platinum atoms in 1 mole of platinum).

$\frac {6.022 \times 10^{23} \ atoms \ Pt}{ 1 \ mol \ Pt}$

We are converting 4.99 ×10²¹ atoms of Pt to moles of Pt, so we multiply by this value.

$4.99 \times 10^{21} \ atoms \ Pt *\frac {6.022 \times 10^{23} \ atoms \ Pt}{ 1 \ mol \ Pt}$

Flip the ratio so the units of atoms of platinum cancel.

$4.99 \times 10^{21} \ atoms \ Pt *\frac { 1 \ mol \ Pt}{6.022 \times 10^{23} \ atoms \ Pt}$

$4.99 \times 10^{21} *\frac { 1 \ mol \ Pt}{6.022 \times 10^{23}}$

$\frac { 4.99 \times 10^{21}}{6.022 \times 10^{23}} \ mol \ Pt$

Divide.

$0.008286283627 \ mol \ Pt$

<h3>2. Convert Moles to Grams </h3>

Next, we convert moles to grams using the molar mass. This is the mass of 1 mole of a substance. This is found on the Periodic Table because it is equivalent to the atomic mass, but the units are grams per mole instead of atomic mass units. Look up platinum's molar mass.

Pt: 195.08 g/mol

Set up another ratio using this new information (195.08 grams of Pt in 1 mole of Pt).

$\frac {195.08 \ g \ Pt}{ 1 \ mol \ Pt}$

Multiply by the number of moles we just calculated.

$0.008286283627 \ mol \ Pt*\frac {195.08 \ g \ Pt}{ 1 \ mol \ Pt}$

The units of moles of platinum cancel.

$0.008286283627*\frac {195.08 \ g \ Pt}{ 1 }$

$0.008286283627* {195.08 \ g \ Pt}$

$1.61648821\ g \ Pt$

<h3>3. Round</h3>

The original measurement of atoms ( 4.99 ×10²¹ ) has 3 significant figures, so our answer must have the same. For the number we found, that is the hundredth place. The 6 in the thousandth place tells us to round the 1 up to a 2.

$1.62 \ g \ Pt$

There are approximately <u>1.62 grams of platinum</u> in 4.99 ×10²¹ atoms of platinum.

8 0

3 years ago

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.8

Lubov Fominskaja [6]

Complete Question

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?

Answer:

The pK_a value is $pK_a =7.82$