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Mekhanik [1.2K]
2 years ago
15

The volume of a single strontium atom is 4.15×10-23 cm3. What is the volume of a strontium atom in microliters

Chemistry
1 answer:
Ivenika [448]2 years ago
7 0

Answer:-  4.15*10^-^2^0\mu L

Solution:- It is a volume unit conversion problem where we are asked to convert the volume from cm^3 to microliters.

We know that:

1cm^3 = 1 mL

1mL=10^-^3L

and, 1L=10^6\mu L

Let's use these conversions factors for the desired conversion using dimensional as:

4.15*10^-^2^3cm^3(\frac{1mL}{1cm^3})(\frac{10^-^3L}{1mL})(\frac{10^6\mu L}{1L})

= 4.15*10^-^2^0\mu L

So, the answer is  4.15*10^-^2^0\mu L .

You might be interested in
How many grams of caf2 would be needed to produce 8.41×10-1 moles of f2?
geniusboy [140]
Answer: 65.7 grams

Explanation:

1) ratio

Since 1 mole of CaF2 contains 1 mol of F2, the ratio is:

1 mol CaF2 : 1 mol F2

2) So, to produce 8.41 * 10^ -1` mol of F2 you need the same number of moles of CaF2.

3) use the formula:

mass in grams = molar mass * number of moles

molar mass of CaF2 = 40.1 g/mol + 2 * 19.0 g/mol = 78.1 g/mol

mass in grams = 78.1 g/mol * 8.41 * 10^ -1 mol = 65.7 grams
5 0
2 years ago
What is the molality of a solution in which 3.0 moles of NaCl is dissolved in 1.5 Kg of water?
Nesterboy [21]
Density H2O = 1g/cm³
1,5 kg H2O = 1500g = 1500cm³             (1dm³ = 1000cm³)

3moles of NaCl-----in---------1500cm³ H2O
x moles of NaCl ----in--------1000cm³ H2O
x = 2moles of NaCl

answer: 2 mol/dm³
5 0
2 years ago
The reaction SO2(g)+2H2S(g)←→3S(s)+2H2O(g) is the basis of a suggested method for removal of SO2 from power-plant stack gases. T
kifflom [539]

Answer : The equilibrium SO_2 pressure is, 3.93\times 10^{-5}torr

Explanation :

The given balanced chemical reaction is,

SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)

First we have to calculate the standard free energy of reaction (\Delta G^o).

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{S(s)}\times \Delta G_f^0_{(S(s))}+n_{H_2O(g)}\times \Delta G_f^0_{(H_2O(g))}]-[n_{SO_2(g)}\times \Delta G_f^0_{(SO_2(g))}+n_{H_2S(g)}\times \Delta G_f^0_{(H_2S(g))}]

where,

\Delta G^o = standard free energy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:

\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (-228.57kJ/mol)]-[1mole\times (-300.4kJ/mol)+2mole\times (-33.01kJ/mol)]

\Delta G^o=-90.72kJ/mol

Now we have to calculate the value of K_p

\Delta G^o=-RT\ln K_p

where,

\Delta G_^o =  standard Gibbs free energy  = -90.72 kJ/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

K_p = equilibrium constant  = ?

Now put all the given values in this expression, we get:

-90.72kJ/mol=-(8.314J/mol.K)\times (298K) \ln K_p

K_p=7.98\times 10^{15}

Now we have to calculate the value of K_p.

The given balanced chemical reaction is,

SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)

The expression for equilibrium constant will be :

K_p=\frac{(p_{H_2O})^2}{(p_{H_2S})^2\times (p_{SO_2})}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Let the equilibrium SO_2 pressure be, x

Pressure of SO_2 = Pressure of H_2S = x

Now put all the given values in this expression, we get

7.98\times 10^{15}=\frac{(22)^2}{(x)^2\times (x)}

x=3.93\times 10^{-5}torr

Thus, the equilibrium SO_2 pressure is, 3.93\times 10^{-5}torr

4 0
2 years ago
At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2 + O2(g) → 2SO3(g) At equilibrium, the partia
lapo4ka [179]

Answer:

The partial pressure of SO₃ is 82.0 atm

Explanation:

The equilibrium constant Kp is equal to <em>the equilibrium pressure of the gaseous products raised to the power of their stoichiometric coefficients divided by the equilibrium pressure of the gaseous reactants raised to the power of their stoichiometric coefficients</em>.

For the reaction,

2 SO₂(g) + O₂(g) → 2 SO₃(g)

Kp = 0.345 = \frac{(pSO_{3})^{2} }{(pSO_{2})^{2} \times pO_{2} }\\pSO_{3} = \sqrt[]{0.345 \times (pSO_{2})^{2} \times pO_{2} } \\pSO_{3} = \sqrt[]{0.345 \times (35.0)^{2} \times 15.9 } \\pSO_{3} = 82.0 atm

4 0
2 years ago
The pOH of a solution is 6.0. Which statement is correct? Use p O H equals negative logarithm StartBracket upper O upper H super
Katen [24]

Answer:

The pH of the solution is 8.

Explanation:

To which options are correct, let us determine the concentration of the hydroxide ion, [OH-] and the pH of the solution. This is illustrated below:

1. The concentration of the hydroxide ion, [OH-] can be obtained as follow:

pOH = –Log [OH-]

pOH = 6

6 = –Log [OH-]

–6 = Log [OH-]

[OH-] = Antilog (–6)

[OH-] = 1x10^–6 mol/L

2. The pH of the solution can be obtained as follow:

pH + pOH = 14

pOH = 6

pH + 6 = 14

pH = 14 – 6

pH = 8.

From the calculations made above,

[OH-] = 1x10^–6 mol/L

pH = 8.

Therefore, the correct answer is:

The pH of the solution is 8

3 0
2 years ago
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