Myra wants to measure the pH of a sample of drinking water as accurately as possible. What method should she use?

How to balance the combustion equation: ___C5H12 + ___O2 ⟶ ___CO2 + ___H2O

photoshop1234 [79]

C5H12 + 8O2 ⟶ 5CO2 + 6H2O

5 0

3 years ago

1. To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 4.8-L bulb, then fil

Delvig [45]

Answer:

1) The diometic gas is N2 (molar mass 28 g/mol)

2) The partialpressure of oxygen is 316.6 mmHg

Explanation:

Step 1: Data given

Volume = 4.8 L

pressure = 1.60 atm

temperature = 30.0°C

Difference in mass after weighting again = 8.7 grams

Step 2:

PV = nRT

⇒ with P = the pressure of the gas = 1.60 atm

⇒ with V = the volume of the gas = 4.8 L

⇒ with n = the number of moles = mass/molar mass

⇒ with R = the gas constant = 0.08206 L*atm/k*mol

⇒ with T = the temperature = 30.0 °C = 303 K

(1.60 atm) (4.8L) = (n)*(0.08206)*(303 K)

n = (1.60 * 4.8) / ( 0.08206*303)

n = 0.30888 mol

Step 3: Calculate molar mass

Molar mass = mass / moles

Molar mass = 8.7 grams / 0.3089 moles

Molar mass ≈ 28 g/mol

The diometic gas is N2

2) What is the partial pressure of oxygen in the mixture if the total pressure is 545mmHg ?

Step 1: Calculate mass of nitrogen

Let's assume a 100 gram sample. This means 38.8 grams is nitrogen

Step2: Calculate moles of N2

Moles N2 = mass N2 / molar mass N2

Moles N2 = 38.8 grams / 28 .02 grams

Moles N2 = 1.38 moles

Step 3: Calculate moles of O2

Moles O2 = (100 - 38.8)/ 32 g/mol

Moles O2 = 1.9125 moles O2

Step 4: Calculate molefraction of oxygen

Molefraction O2 = moles of component/total moles in mixture

=1.9125/(1.9125 + 1.38 moles)

=0.581

Step 5: Calculate the partial pressure of oxygen

PO2 =molefraction O2 * Ptotal

=0.581 * 545mmHg

=316.6 mmHg

The partialpressure of oxygen is 316.6 mmHg

7 0

3 years ago

A cylinder with a radius of 5.6 cm and a height of 10.7 cm has what volume? (You can look up the formula for the volume of a cyl

Keith_Richards [23]

Answer:

$V=1054.2cm^{3}$

Explanation:

1. First take the cylinder volume formula:

$V=\pi.r^{2}.h$

2. Then take the values for the radius and the height of the cylinder and replace them into the formula:

$V=\pi*(5.6cm)^{2}*(10.7cm)$

3. Solve the equation:

$V=\pi*(31.36cm^{2})*(10.7cm)$

$V=\pi*335.5cm^{3}$

$V=1054.2cm^{3}$

6 0

4 years ago

One solution has a formula C (n) H (2n) O (n) If this material weighs 288 grams, dissolves in weight 90 grams, the solution will

saw5 [17]

Explanation:

The given data is as follows.

Boiling point of water ( $T^{o}_{b}) = 100^{o}C$ = (100 + 273) K = 323 K,

Boiling point of solution ( $T_{b}) = 101.24^{o}C$ = (101.24 + 273) K = 374.24 K

Hence, change in temperature will be calculated as follows.

$\Delta T_{b} = (T_{b} - T^{o}_{b})$

= 374.24 K - 323 K

= 1.24 K

As molality is defined as the moles of solute present in kg of solvent.

Molality = $\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}$

Let molar mass of the solute is x grams.

Therefore, Molality = $\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}$

m = $\frac{288 g \times 1000}{x g \times 90}$

= $\frac{3200}{x}$

As, $\Delta T_{b} = k_{b} \times molality$

$1.24 = 0.512 ^{o}C/m \times \frac{3200}{x}$

x = $\frac{0.512 ^{o}C/m \times 3200}{1.24}$

= 1321.29 g

This means that the molar mass of the given compound is 1321.29 g.

It is given that molecular formula is $C_{n}H_{2n}O_{n}$ .

As, its empirical formula is $CH_{2}O$ and mass is 30 g/mol. Hence, calculate the value of n as follows.

n = $\frac{\text{Molecular mass}}{\text{Empirical mass}}$

= $\frac{1321.29 g}{30 g/mol}$

= 44 mol

Thus, we can conclude that the formula of given material is $C_{44}H_{88}O_{44}$ .

4 0

3 years ago

A mixture of gaseous reactants is put into a cylinder, where a chemical reaction turns them into gaseous products. The cylinder

mote1985 [20]

Answer:

a) The reaction is exothermic

b) The temperature of the water goes up

c) The piston move in

d) Work is done on the gas mixture

e) The amount of work done on the gas mixture is given as

W = PΔV

where

W is the work done on the gas mixture

P is the pressure within the cylinder after the reaction

ΔV is the change (a decrease) in volume of the gas mixture

Explanation:

a) Since the chemical reaction is known to release energy into the surrounding, then it is an exothermic reaction.

b) Sine the reaction releases energy into the water, the energy of the water will go up, causing the temperature of the water to go up.

c) The piston will move in, since it was determined from data that the piston does work on the system. When the piston moves in work is done on the system.

d) Work is done on the gas mixture, because the piston moves in, compressing the gas.

e) From basic thermodynamics, we know that when work is done a system, then it means the system is compressed, and the equation for the work done on a system is given as

W = PΔV

where

P is the pressure withing the cylinder

ΔV is the change in the volume of the system. The volume of the system decreases, and hence the work done will be negative.

5 0

3 years ago