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3 years ago

12.5 mL of 0.280 M HNO3 and 5.0 mL of 0.920 M KOH are mixed. Is the resulting solution acidic, basic or neutral?

Chemistry

1 answer:

svlad2 [7]3 years ago

8 0

Answer:

The resulting solution is basic.

Explanation:

The reaction that takes place is:

HNO₃ + KOH → KNO₃ + H₂O

First we <u>calculate the added moles of HNO₃ and KOH</u>:

HNO₃ ⇒ 12.5 mL * 0.280 M = 3.5 mmol HNO₃

KOH ⇒ 5.0 mL * 0.920 M = 4.6 mmol KOH

As <em>there are more KOH moles than HNO₃,</em> the resulting solution is basic.

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As the temperature of one of the four liquids increases, the vapor pressure will

s2008m [1.1K]

Answer:

a) increase exponentially.

Explanation:

The vapor pressure is depend only on temperature.

The vapor pressure of liquid does not depend upon amount of liquid. For example whether the liquid is 50 g or 30 g its vapor pressure will remain same according to the temperature.

The temperature and vapor pressure have exponential relationship. As the temperature of liquid increases its vapor pressure also goes to increase. When the temperature of liquid goes to decrease its vapor pressure also decreases.

The change in vapor pressure of substance when temperature changes is given as,

ln P₂/P₁ = ΔH(va)/R (1/T₁ - 1/T₂)

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3 years ago

Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow

andreyandreev [35.5K]

Explanation:

(a) The given data is as follows.

Pressure on top ( $P_{o}$ ) = 140 bar = $1.4 \times 10^{7} Pa$ (as 1 bar = $10^{5}$ )

Temperature = $15^{o}C$ = (15 + 273) K = 288 K

Density of gas = $\frac{PM}{ZRT}$

$\frac{dP}{dZ} = \rho \times g$

$\frac{dP}{dZ} = \int \frac{PM}{ZRT}$

$\int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ$

$ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}$

= 0.4548

$P_{1} = P_{o} \times e^{0.4548}$

= $1.4 \times 10^{7} Pa \times 1.5797$

= $2.206 \times 10^{7} Pa$

Hence, pressure at the natural gas-oil interface is $2.206 \times 10^{7} Pa$ .

(b) At the bottom of the tank,

$P_{2} = P_{1} + \rho \times g \times h$

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

= $309.8 \times 10^{5} Pa$

= 309.8 bar

Hence, at the bottom of the well at $15^{o}C$ pressure is 309.8 bar.

6 0

3 years ago

What volume would 3.01•1023 molecules of oxygen gas occupy at STP?

Sliva [168]

First, find moles of oxygen gas: (3.01 x10^23 molec.)/(6.02 x10^23) =0.5mol O2

Second, multiply moles by the standard molar volume of a gas at STP:(0.5mol)(22.4L) = 11.2L O2

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3 years ago

If the distance between two objects is decreased to - of the original

Charra [1.4K]

Answer:

The new force will be \frac{1}{100} of the original force.

Explanation:

In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.

That said, let's say that our equation for the initial force is:

$F = G\frac{m_1m_2}{R^2}The problem states that the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R$

And the force at this distance would be written in terms of the same equation:

$F_2 = G\frac{m_1m_2}{R_2^2}$

Find the ratio between the final and the initial force:

$\frac{F_2}{F} = \frac{G\frac{m_1m_2}{R_2^2}}{G\frac{m_1m_2}{R^2}}$

Substitute the value for the final distance in terms of the initial distance:

$\frac{F_2}{F} = \frac{G\frac{m_1m_2}{(\frac{R}{10})^2}}{G\frac{m_1m_2}{R^2}}$

Simplify:

$\frac{F_2}{F} = \frac{\frac{1}{100R^2}}{\frac{1}{R^2}}=\frac{1}{100}$

This means the new force will be \frac{1}{100} of the original force.

8 0

3 years ago

The combustion of 0.1240 kg of propane in the presence of excess oxygen produces 0.3110 kg of carbon dioxide. What is the limiti

nadezda [96]

Answer:

The limiting reactant is the propane gas, C₃H₈ while the percentage yield is 83.77%

Explanation:

Here we have

Propane gas with molecular formula C₃H₈, molar mass = 44.1 g/mol combining with O₂ as follows

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Therefore, 1 mole of C₃H₈ combines with 5 moles of O₂ to produce 3 moles CO₂ and 4 moles of H₂O

Mass of propane = 0.1240 kg = 124.0 g

Number of moles of propane = mass of propane/(molar mass of propane)

The number of moles of propane = 124/44.1 = 2.812 moles

The molar mass of CO₂ = 44.01 g/mol

Mass of CO₂ = 0.3110 kg = 311.0 g

Therefore, number of moles of CO₂ = mass of CO₂/(molar mass of CO₂)

The number of moles of CO₂ = 311.0 kg/ 44.01 g/mol = 7.067 moles

Therefore, since 1 mole of propane produces 3 moles of CO₂, 2.812 moles of propane will produce 3 × 2.812 moles or 8.44 moles of CO₂

Therefore;

The limiting reactant is the propane gas, C₃H₈, since the oxygen is in excess

Hence

$The \ percentage \ yield = \frac{Actual \, yield}{Theoretical \, yield} \times 100 = \frac{7.067}{8.44} \times 100 = 83.77 \%$

The percentage yield = 83.77%.

7 0

3 years ago