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3 years ago

What is the molar mass to Ca3(Po4)2

1 answer:

4 0

P (phosphorous ) = 30.97 g/mol * 2 = 61.94 g/mol.
O ( oxygen ) = 16.00 g/mol * 8 = 128.00 g/mol.
-----------------------------------------------------------+
Calcium Phosphate Ca3(PO4)2 = 310.18 g/mol.

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Who conducted experiments to determine the quantity of charge carried by an electron?

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I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

You must round to two significant figures: 290 ml

Problem 4

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

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3 years ago

A gas at 928 kpa, 129 C occupies a volume of 569 L. Calculate the volume at 319 kpa and 32 C.

lisabon 2012 [21]

Answer:

1255.4L

Explanation:

Given parameters:

P₁ = 928kpa

T₁ = 129°C

V₁ = 569L

P₂ = 319kpa

T₂ = 32°C

Unknown:

V₂ = ?

Solution:

The combined gas law application to this problem can help us solve it. It is mathematically expressed as;

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

P, V and T are pressure, volume and temperature

where 1 and 2 are initial and final states.

Now,

take the units to the appropriate ones;

kpa to atm, °C to K

P₂ = 319kpa in atm gives 3.15atm

P₁ = 928kpa gives 9.16atm

T₂ = 32°C gives 273 + 32 = 305K

T₁ = 129°C gives 129 + 273 = 402K

Input the values in the equation and solve for V₂;

$\frac{9.16 x 569}{402} = \frac{3.15 x V_{2} }{305}$

V₂ = 1255.4L

4 0

3 years ago