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3 years ago

How do you find the volume of a liquid

1 answer:

7 0

Today they will practice measuring different liquids. They will use a container called a graduated cylinder to measure liquids. Graduated cylinders have numbers on the side that help you determine the volume. Volume is measured in units called liters or fractions of liters called milliliters (ml).

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Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simulta

Brut [27]

Answer:

The molar mass of unknown gas is 145.82 g/mol.

Explanation:

Volume of oxygen gas effused under time t = 8.24 mL

Effusion rate of oxygen gas = $R=\frac{8.24 mL}{t}$

Molar mass of oxygen gas = 32 g/mol

Volume of unknown gas effused under time t = 3.86 mL

Effusion rate of unknown gas = $R'=\frac{3.86 mL}{t}$

Molar mass of unknown gas = M

Graham's Law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

$\frac{R}{R'}=\sqrt{\frac{M}{32 g/mol}}$

$\frac{\frac{8.24 mL}{t}}{\frac{3.86 mL}{t}}=\sqrt{\frac{M}{32 g/mol}}$

$M=\frac{32 g/mol\times 8.24 \times 8.24}{3.86\times 3.86}=145.82 g/mol$

4 0

3 years ago

When you pour 125 grams of solute into 1.25 Liters of 35 grams of crystal sink to the bottom and do not dissolve. What is the sa

Sati [7]

Answer:

72 g/L

Explanation:

The dissolved amount of solute is the difference between the amount you have poured and the amount that precipitated:

125 g - 35 g = 90 g

Thus, 90 grams of solute were dissolved in 1.25 liters. The saturation point is the ratio between the grams dissolved and the volume in liters:

saturation point = 90 g/1.25 L = 72 g/L

4 0

3 years ago

The density of mercury is 13.6 g/mL what is the density in lbs/L ( 1 lb hint =0.454 kg )

Vesna [10]

Answer:

$30.0\frac{lb}{L}$

Explanation:

Hello,

In this case, since 454 g are equivalent to 1 pound and 1000 millilitres are equivalent to 1 liter, the required density is computed below by applying the corresponding conversion factor:

$=13.6\frac{g}{mL} *\frac{1lb}{454g} *\frac{1000mL}{1L} \\\\=30.0\frac{lb}{L}$

Regards.

6 0

4 years ago

How many moles of lithium are present in a sample that contains 2.45x10^87 formula units of Li2SO4?

ddd [48]

The answer is: 8.14·10⁶³ moles of lithium are present.

N(Li₂SO₄) = 2.45·10⁸⁷; number of formula units of lithium sulfate.

n(Li₂SO₄) = N(Li₂SO₄) ÷ Na.

n(Li₂SO₄) = 2.45·10⁸⁷ ÷ 6.022·10²³ 1/mol.

n(Li₂SO₄) = 4.07·10⁶³ mol; amount of lithium sulfate

In one molecule of lithium sulfate, there are two atoms of lithium.

n(Li₂SO₄) : n(Li) = 1 : 2.

n(Li) = 2 · 4.07·10⁶³ mol.

n(Li) = 8.14·10⁶³ mol; amount of lithium atoms.

3 0

3 years ago

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl−+I−→OI−+Cl− T

motikmotik

Answer :

(a) The rate law for the reaction is:

$\text{Rate}=k[OCl^-]^1[I^-]^1$

(b) The value of rate constant is, $60.4M^{-1}s^{-1}$

(c) rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$OCl^-+I^-\rightarrow OI^-+Cl^-$

Rate law expression for the reaction:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

where,

a = order with respect to $OCl^-$

b = order with respect to $I^-$

Expression for rate law for first observation:

$1.36\times 10^{-4}=k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(1)

Expression for rate law for second observation:

$2.72\times 10^{-4}=k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(2)

Expression for rate law for third observation:

$2.72\times 10^{-4}=k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}\\\\2=2^a\\a=1$

Dividing 1 from 3, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b}\\\\2=2^b\\b=1$

Thus, the rate law becomes:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

a = 1 and b = 1

$\text{Rate}=k[OCl^-]^1[I^-]^1$

Now, calculating the value of 'k' (rate constant) by using any expression.

$1.36\times 10^{-4}=k(1.5\times 10^{-3})(1.5\times 10^{-3})$

$k=60.4M^{-1}s^{-1}$

Now we have to calculate the rate for a reaction when concentration of $OCl^-$ and $I^-$ is $1.8\times 10^{-3}M$ and $6.0\times 10^{-4}M$ respectively.

$\text{Rate}=k[OCl^-][I^-]$

$\text{Rate}=(60.4M^{-1}s^{-1})\times (1.8\times 10^{-3}M)(6.0\times 10^{-4}M)$

$\text{Rate}=6.52\times 10^{-5}Ms^{-1}$

Therefore, the rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

8 0

3 years ago