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Ivan
3 years ago
6

A duck migrates South for the winter about 3000 miles each year. It takes them about 65 hours to travel this distance. What is t

he velocity of the duck?
Chemistry
1 answer:
olchik [2.2K]3 years ago
7 0

Answer:

46.15

Explanation:

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A. Only a small amount (less than 0.01%) of the enol form of diethyl malonate is present at equilibrium. Write a structural form
Karo-lina-s [1.5K]

Answer:

Please see attachment

Explanation:

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6 0
3 years ago
The solubility product for barium sulfate is 1.1 × 10–10. calculate the molar solubility of barium sulfate.
Triss [41]
When it comes to equilibrium reactions in chemistry, there are a lot of equilibrium constants that can be used. In the case of solubility, the appropriate one to use is the equilibrium constant of solubility product denotes as Ksp. This is the concentration of products raised to their coefficients. For example,

 cC ⇔ aA + bB 

Ksp = {[A^a][B^b]}

Now, for the this problem, the reaction is

BaSO₄ ⇔ Ba²⁺ + SO₄²⁻

The reaction is already balanced. Since we don't know the value of Ba²⁺ and SO₄²⁻, let's denote this at x.

1.1 × 10⁻¹⁰ = [x][x] =[x²]
[x] = [Ba²⁺] = [SO₄²⁻] = [BaSO₄] = 1.049 × 10⁻⁵ M
8 0
3 years ago
8.45 cm - 3.55 cm = ?
lyudmila [28]

Answer:

4.9cm or 4.90cm

Explanation:

Use column method and align the numbers. Make sure you add the decimal point before you write the answer, otherwise the value would be wrong.

6 0
3 years ago
Read 2 more answers
Give the systematic name for this coordination compound. k3 cr(cn)6
Helen [10]

Potassium hexacyanochromate(III)

A systematic name is a name given in a systematic way to a chemical substance, out of a definite collection.

The systematic name for the coordination compound k3 cr(cn)6 is Potassium hexacyanochromate(III). This compound contains potassium (k3), six cn which is called hexacyano, and cr (chromium).

 






8 0
3 years ago
The table shows the amount of radioactive element remaining in a sample over a period of time.
Lunna [17]

Answer:

1. The half-life is 22 years.

2. 132 years

Explanation:

1. Determination of the the half-life.

The half-life of an element is the time taken for half the element to decay.

From the table given above, the original amount of the element 45 g. If we divide 45 by 2, we'll have 22.5 g as half the original amount of element.

Now, the time taken to obtain 22.5 g as shown from the table is 22 years.

Thus, the half-life the element is 22 years.

2. Determination of the time.

Original amount (N₀) = 308 g

Amount remaining (N) = 4.8125 g

Time (t) =?

Next, we shall the number of half-lives that has elapsed. This can be obtained as follow:

Original amount (N₀) = 308 g

Amount remaining (N) = 4.8125 g

Number of half-lives (n) =?

N = 1/2ⁿ × N₀

4.8125 = 1/2ⁿ × 308

Cross multiply

4.8125 × 2ⁿ = 308

Divide both side by 4.8125

2ⁿ = 308 / 4.8125

2ⁿ = 64

Express 64 in index form with 2 as the base.

2ⁿ = 2⁶

n = 6

Thus, 6 half-lives has elapsed.

Finally, we shall determine the time. This can be obtained as follow:

Number of half-lives (n) = 6

Half-life (t½) = 22 years

Time (t) =?

n = t / t½

6= t / 22 years

Cross multiply

t = 6 ×22

t = 132 years.

Thus, the time taken is 132 years.

6 0
3 years ago
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