How many bonds exist in this molecule?

When 15.6 g of fluorite (caf2) reacts with excess sulfuric acid, hydrogen fluoride gas is collected at 709 torr and 25.5Â°c. sol

lianna [129]

The reaction equation is:

CaF₂ + H₂SO₄ → 2HF + CaSO₄

The molar ratio between fluorite and hydrogen fluoride is 1 : 1.

The moles of fluorite supplied are:

Moles = 15.6 / 78.07
Moles = 0.200

The moles of hydrogen fluoride produced will be 0.2.

Now, we may use the ideal gas equation to determine the temperature:

PV = nRT
T = PV/nR

T = (899 * 7.4) / (0.2 * 62.36)

T = 533.40 K

The temperature will be 260.25 °C

5 0

3 years ago

The reaction of hydrogen gas with oxygen gas is what type of reaction precipitation single replacement

Oksi-84 [34.3K]

The reaction of hydrogen gas (H2) with oxygen gas (O2) is a COMBINATION or SYNTHESIS reaction, because multiple substances combine to form fewer substances. Here, the two gases form one substance, water (H2O):
2H2 + O2 -- > 2H2O

5 0

3 years ago

State the rank in the Brainly apk

Ket [755]

Explanation:

nah biaar seneng, tuh ku jawab:c

3 0

3 years ago

Arrhenius base definition

kompoz [17]

Arrhenius base is a substance that , when dissolved in an aqueous solution , increase the concentration of hydroxide (OH) ion in the solution

I hope that's help !

6 0

3 years ago

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:

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

3 0

3 years ago