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Leokris [45]
3 years ago
14

The object whose motion is shown on the following graph is

Chemistry
1 answer:
Tanzania [10]3 years ago
8 0
I don’t understand what your asking please explain more
You might be interested in
Please help me on this!!
Fudgin [204]

Answer:

OK

Explanation:

OK

7 0
3 years ago
(2 pts) The solubility of InF3 is 4.0 x 10-2 g/100 mL. a) What is the Ksp? Include the chemical equation and Ksp expression. MW
Aleonysh [2.5K]

Answer:

a) Ksp = 7.9x10⁻¹⁰

b) Solubility is 6.31x10⁻⁶M

Explanation:

a) InF₃ in water produce:

InF₃ ⇄ In⁺³ + 3F⁻

And Ksp is defined as:

Ksp = [In⁺³] [F⁻]³

4.0x10⁻²g / 100mL of InF₃ are:

4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = <em>2.3x10⁻³M  InF₃. </em>Thus:

[In⁺³] = 2.3x10⁻³M  InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M  In⁺³

[F⁻] = 2.3x10⁻³M  InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻

Replacing these values in Ksp formula:

Ksp = [2.3x10⁻³M  In⁺³] × [7.0x10⁻³M F⁻]³ = <em>7.9x10⁻¹⁰</em>

<em></em>

b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:

7.9x10⁻¹⁰ = [x] [0.05 + 3x]³

Where x are moles of In⁺³ produced from solid InF₃ and 3x are moles of F⁻ produced from the same source. That means x is solubility in mol / L

Solving from x:

x = -0.018 → False solution, there is no negative concentrations.

x = 6.31x10⁻⁶M → Right answer.

Thus, <em>solubility is 6.31x10⁻⁶M</em>

3 0
3 years ago
the density of glycerin is 1.26 g/cm3. How many pounds/foot3 is this ? use the conversion rates of 454g/1 pound and 28,317cm3/1f
lyudmila [28]

Answer:

78.6 lb/ft³  

Step-by-step explanation:

Let's do this in steps.

1. Convert grams to pounds

D = (1.26 g/1 cm³) × (1 lb/454 g)

   = 2.775 × 10⁻³ lb/cm³

2. Convert cubic centimetres to cubic feet

D = (2.775 × 10⁻³ lb/1 cm³) × (28 317cm³/1 ft³)

   = 78.6 lb/ft³

6 0
2 years ago
A 1.52 g sample of KCIO3 is reacted according to the balanced equation below. How many liters of O2 is produced at a pressure of
ipn [44]

Answer:

0.486 L

Explanation:

Step 1: Write the balanced reaction

2 KCIO₃(s) ⇒ 2 KCI (s) + 3 O₂(g)

Step 2: Calculate the moles corresponding to 1.52 g of KCIO₃

The molar mass of KCIO₃ is 122.55 g/mol.

1.52 g × 1 mol/122.55 g = 0.0124 mol

Step 3: Calculate the moles of O₂ produced from 0.0124 moles of KCIO₃

The molar ratio of KCIO₃ to O₂ is 2:3. The moles of O₂ produced are 3/2 × 0.0124 mol = 0.0186 mol

Step 4: Calculate the volume corresponding to 0.0186 moles of O₂

0.0186 moles of O₂ are at 37 °C (310 K) and 0.974 atm. We can calculate the volume of oxygen using the ideal gas equation.

P × V = n × R × T

V = n × R × T/P

V = 0.0186 mol × (0.0821 atm.L/mol.K) × 310 K/0.974 atm = 0.486 L

7 0
2 years ago
Astatine-210 has a half-life of 8.08 days. What fraction of a sample of astatine-210 is left unchanged after 16.16 days?
Flauer [41]

Answer:

0

Explanation:

Given parameters:

Half-life  = 8.08days

Unknown:

What fraction is left unchanged after 16.16days = ?

Solution:

The half - life of a substance is the time taken for the half of a radioactive material to decay to half.

 

  Day 0           Day 8.08         Day 16.16

   100%                 50%                 0%        Parent

    0%                     50%                100%     Daughter

After 16.16 days, non of the original sample will remain unchanged.

6 0
3 years ago
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