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3 years ago

A sample of 1.00 moles of oxygen at 50°C and 98.6 kPa, occupies what volume?

Chemistry

1 answer:

enyata [817]3 years ago

5 0

Answer:

27.22 dm³

Explanation:

Given parameters:

number of moles = 1 mole

temperature= 50°C, in K gives 50+ 273 = 323K

Pressure= 98.6kpa in ATM, gives 0.973 ATM

Solution:

Since the unknown is the volume of gas, applying the ideal gas law will be appropriate in solving this problem.

The ideal gas law is mathematically expressed as,

Pv=nRT

where P is the pressure of the gas

V is the volume

n is the number of moles

R is the gas constant

T is the temperature

Input the parameters and solve for V,

0.973 x V = 1 x 0.082 x 323

V= 27.22 dm³

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When the volume of a gas is

Montano1993 [528]

Answer:

$\boxed {\boxed {\sf 232.9 \textdegree C}}$

Explanation:

This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:

$\frac {V_1}{T_1}= \frac{V_2}{T_2}$

The volume of the gas starts at 250 milliliters and the temperature is 137 °C.

$\frac{250 \ mL}{137 \textdegree C}= \frac{V_2}{T_2}$

The volume of the gas is increased to 425 milliliters, but the temperature is unknown.

$\frac{250 \ mL}{137 \textdegree C}= \frac{425 \ mL}{T_2}$

We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.

$250 \ mL * T_2 = 137 \textdegree C * 425 \ mL$

Now the variable is being multiplied by 250 milliliters. The inverse of multiplication is division. Divide both sides of the equation by 250 mL.

$\frac{250 \ mL * T_2}{250 \ mL}=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}$

$T_2=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}$

The units of milliliters (mL) cancel.

$T_2=\frac{ 137 \textdegree C * 425 }{250 }$

$T_2= \frac{58225}{250} \textdegree C$

$T_2=232.9 \textdegree C$

The temperature changes to <u>232.9 degrees Celsius.</u>

3 0

3 years ago

In a living organism, what is an organ? functional unit, or building block, of all organisms; smallest unit that can carry on th

ra1l [238]

Answer: D. a structure that provides a specific service within a cell

Explanation:

7 0

2 years ago

8. The time period of artificial satellite in a circular orbit of radius R is T. The radius of the orbit in which time period is

Elena-2011 [213]

Explanation:

It is given that,

The time period of artificial satellite in a circular orbit of radius R is T. The relation between the time period and the radius is given by :

$T^2\propto R^3$

The radius of the orbit in which time period is 8T is R'. So, the relation is given by :

$(\dfrac{T}{T'})^2=(\dfrac{R}{R'})^3$

$(\dfrac{T}{8T})^2=(\dfrac{R}{R'})^3$

$\dfrac{1}{64}=(\dfrac{R}{R'})^3$

$R'=4\times R$

So, the radius of the orbit in which time period is 8T is 4R. Hence, this is the required solution.

4 0

3 years ago

Write ionic equations for any three of the following:

worty [1.4K]

The ionic equation is:

H⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + NO₃⁻(aq) + H₂O(l)

HNO₃(aq) + NaOH(aq) → NaNO₃(aq) + H2O(l)

The required dissociations are

HNO₃(aq) → H⁺(aq) + NO₃⁻(aq)

NaOH(aq) → Na⁺(aq) + OH⁻(aq)

NaNO₃(aq) → Na⁺(aq) + NO₃⁻(aq)

So, the ionic equation is

H⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + NO₃⁻(aq) + H₂O(l)

The ionic equation is

H⁺(aq) + Cl⁻(aq) + K⁺(aq) + OH⁻(aq) → K⁺(aq) + Cl⁻(aq) + H₂O(l)

HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)

The required dissociations are

HCl(aq) → H⁺(aq) + Cl⁻(aq)

KOH(aq) → K⁺(aq) + OH⁻(aq)

KCl(aq) → K⁺(aq) + Cl⁻(aq)

So, the ionic equation is

H⁺(aq) + Cl⁻(aq) + K⁺(aq) + OH⁻(aq) → K⁺(aq) + Cl⁻(aq) + H₂O(l)

The ionic equation is

2H⁺(aq) + SO₄²⁻(aq) + Mg²⁺(aq) + 2OH⁻(aq) → Mg²⁺(aq) + SO₄²⁻(aq)

+ 2H₂O(l)

H₂SO₄(aq) + Mg(OH)₂(aq) → MgSO₄(aq) + 2H2O(l)

The required dissociations are

H₂SO₄(aq) → 2H⁺(aq) + SO₄²⁻(aq)

Mg(OH)₂(aq) → Mg²⁺(aq) + 2OH⁻(aq)

MgSO₄(aq) → Mg²⁺(aq) + SO₄²⁻(aq)

So, the ionic equation is

2H⁺(aq) + SO₄²⁻(aq) + Mg²⁺(aq) + 2OH⁻(aq) → Mg²⁺(aq) + SO₄²⁻(aq)

+ 2H₂O(l)

Learn more about ionic equations here:

brainly.com/question/11628165

5 0

3 years ago

An exponent of "2" means that if we double the concentration of the reactant the rate doubles as well Exponents in rate laws are

Karo-lina-s [1.5K]

Answer:

- False.

- True.

Explanation:

Hello, for each statement we state:

- An exponent of "2" means that if we double the concentration of the reactant the rate doubles as well.

FALSE because considering a rate law like:

$-r=kC^2$

The exponent of "2" powers the concentration to the second power, not doubles the rate law, thus, if C is 3, for k=1, r will be -9. On the other hand if the rate is like:

$-r=kC$

The rate will be -3, that is why the rate is not doubled when the "2" in concentration is present.

- Exponents in rate laws are based on the coefficients from the balanced equation.

FALSE because for nonelemental chemical reactions, the exponents do not match with each species' stoichiometric coefficients in the rate law.

- The rate constant, k, takes into account the effect of activation energy and temperature on the reaction.

TRUE, since the Arrhenius equation allows us to prove the effect of the activation energy and the temperature:

$k=Aexp(-\frac{Ea}{RT})$

- Differential rate laws allow us to compare concentration and time.

TRUE as they are given like:

$\frac{1}{\nu _A} \frac{dC_A}{dt} =\frac{1}{\nu _B} \frac{dC_B}{dt} =...$

Best regards.

5 0

3 years ago