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1 year ago

A chemical reaction happens

Chemistry

1 answer:

Art [367]1 year ago

5 0

B I think lol when something changes the state of matter

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A beach has a supply of sand grains composed of calcite, ferromagnesian silicate minerals, and non-ferromagnesian silicate miner

bezimeni [28]

Ferromagnesian silicate minerals (i looked it up)

4 0

1 year ago

in a mixture of helium and chlorine, occupying a volume of 12.8 l at 605.6 mmhg and 21.6 oc, it is found that the partial pressu

rodikova [14]

Answer:

Mass of sample = 8.483 g

Explanation:

Given data;

Volume of mixture = 12.8 L

Pressure = 605.6 mmHg ( 605.6 / 760 = 0.797 atm)

Temperature = 21.6 °C (21.6 + 271.15 = 294.8 K)

Partial pressure of chlorine = 143 mmHg ( 143/760 = 0.19 atm)

Solution:

First of all we will determine the number of moles of mixture.

PV = nRT

n = PV/RT

n = 0.797atm × 12.8L / 0.0821 atm. dm³ mol⁻¹ K⁻¹ ×294.8 K

n = 10.202 / 24.2031

n = 0.422 mol

partial pressure of chlorine is 0.19 atm so mole fraction is,

mole fraction = 0.19/0.797

mole fraction = 0.24

moles of chlorine = 0.24 × 0.422 = 0.1013 mol

moles of helium = moles of mixture - moles of chlorine

moles of helium = 0.422 - 0.1013

moles of helium = 0.3207 mol

Mass of chlorine = moles × molar mass

Mass of chlorine = 0.1013 mol × 71 g/mol

Mass of chlorine = 7.2 g

Mass of helium = moles × molar mass

Mass of helium = 0.3207 mol × 4 g/mol

Mass of helium = 1.283 g

Mass of sample = mass of chlorine + mass of helium

Mass of sample = 7.2 g + 1.283 g

Mass of sample = 8.483 g

5 0

1 year ago

Suppose you measure the absorbance of a yellow dye solution in a 1.00 cm cuvette. The absorbance of the solution at 427 nm is 0.

Dimas [21]

Answer:

The concentration of the solution, $C=7.2992\times 10^{-6} M$

Explanation:

The absorbance of a solution can be calculated by Beer-Lambert's law as:

$A=\varepsilon Cl$

Where,

A is the absorbance of the solution

ɛ is the molar absorption coefficient ( $L.mol^{-1}.cm^{-1}$ )

C is the concentration ( $mol^{-1}.L^{-1}$ )

l is the path length of the cell in which sample is taken (cm)

Given,

A = 0.20

ɛ = 27400 $M^{-1}.cm^{-1}$

l = 1 cm

Applying in the above formula for the calculation of concentration as:

$A=\varepsilon Cl$

$0.20= 27400\times C\times 1$

$C = \frac{0.20}{27400\times 1} M$

So , concentration is:

$C=7.2992\times 10^{-6} M$

4 0

1 year ago

A 3.5 M KNO3 solution contains 3.5 miles of KNO3 dissolved in A) 1.0 liter of solution B) 1.0 liter of solvent C) 3.5 liters of

Vladimir [108]

N=3.5 mol
c=3.5 mol/L

n=cv

v=n/c

v=3.5/3.5=1.0 L

A) 1.0 liter of solution

4 0

1 year ago

A vial containing radioactive selenium-75 has an activity of 3.0 mCi/mL. If 2.6 mCi are required for a leukemia test, how many m

oksian1 [2.3K]

Answer : The $866.66\mu L$ must be administered.

Solution :

As we are given that a vial containing radioactive selenium-75 has an activity of $3.0mCi/mL$ .

As, 3.0 mCi radioactive selenium-75 present in 1 ml

So, 2.6 mCi radioactive selenium-75 present in $\frac{2.6mCi}{3.0mCi}\times 1ml=0.86666ml\times 1000=866.66\mu L$

Conversion :

$(1ml=1000\mu L)$

Therefore, the $866.66\mu L$ must be administered.

4 0

1 year ago