An instrument that is used to measure air pressure is a ?

Consider the following unbalanced chemical equation. C5H12(l) + O2(g) → CO2(g) + H2O(l) If 21.9 grams of pentane (C5H12) are bur

Marina CMI [18]

Answer : The mass of water produced will be 32.78 grams.

Explanation : Given,

Mass of $C_5H_{12}$ = 21.9 g

Molar mass of $C_5H_{12}$ = 72.15 g/mole

Molar mass of $H_2O$ = 18 g/mole

First we have to calculate the moles of $C_5H_{12}$ .

$\text{Moles of }C_5H_{12}=\frac{\text{Mass of }C_5H_{12}}{\text{Molar mass of }C_5H_{12}}=\frac{21.9g}{72.15g/mole}=0.3035moles$

Now we have to calculate the moles of $H_2O$ .

The balanced chemical reaction will be,

$C_5H_{12}(l)+8O_2(g)\rightarrow 5CO_2(g)+6H_2O(l)$

From the balanced reaction we conclude that

As, 1 mole of $C_5H_{12}$ react to give 6 moles of $H_2O$

So, 0.3035 moles of $C_5H_{12}$ react to give $0.3035\times 6=1.821$ moles of $H_2O$

Now we have to calculate the mass of $H_2O$ .

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O$

$\text{Mass of }H_2O=(1.821mole)\times (18g/mole)=32.78g$

Therefore, the mass of water produced will be 32.78 grams.

3 0

2 years ago

A sample of sugar (C12H22O11) contains

Lady_Fox [76]

4 moles of sugar.

Explanation:

A mole is defined as the amount of a substance contained in Avogadro's number of particles 6.02 x 10²³.

1 mole of substance = 6.02 x 10²³. molecules

Given that;

the sample of sugar contains 1.505 x 10²³.molecules

The number of moles in this amount of sugar is 4 moles

Learn more:

Number of moles brainly.com/question/13064292

#learnwithBrainly

5 0

2 years ago

A neutral solution has a pH =

Aleonysh [2.5K]

Answer:

7.

Explanation:

A neutral solution has a pH=7.

A basic solution has a pH>7.

An acidic solution has a pH<7.

3 0

2 years ago

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

2 years ago

Consider a solution containing .100 M fluoride ions and .126M hydrogen fluoride. The concentration of fluoride ions after the ad

S_A_V [24]

Given:

Concentration of Fluoride ions = 0.100 M
Concentration of Hydrogen Fluoride = 0.126 M

Asked: Concentration of fluoride ions after the addition of 5ml of 0.0100 M HCl to 25 mL of the solution

Assume: 50:50 ratio of fluoride ions and HF

12.5ml*0.1mol/L *1L/1000mL + 12.5*0.126mol/L * 1L/1000mL = 2.825x10^-3 moles F-

5ml * 0.01 mol/L *1L/1000mL = 5x10^-5 moles

Assume: Volume additive

Final concentration = 2.825x10^-3 + 5x10^-5 moles/ 30 ml * 1000ml/L =0.0958 M

3 0

2 years ago