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4 years ago

A three-carbon alkene contains two double bonds. Predict its molecular formula? A. C3H4 B. C3H5 C. C3H6

1 answer:

7 0

The correct answer is A. C3H4 is the three-carbon alkene contains two double bonds. Alkenes are a class of organic compounds with carbon-carbon double bond. The structure for C3H4 is:

H2C=C=CH2

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Calcolate resulant force 15N <-[]->20N

Reil [10]

Explanation:

5N right because -15N for left plus +20N for right = +5N

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3 years ago

What element is similar to Baruim

aivan3 [116]

Answer:

calcium ca

Explanation:

We can see here that the only element that is on the same group (column) as Ba (Barium) is Calcium (Ca).

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3 years ago

Read 2 more answers

An experiment requires 42.5 g of sulfuric acid, which is easier to measure by volume than by mass what voliume should be measure

kodGreya [7K]

Using proportions, if x is the unknown volume we have:
1.84 / 1 = 42.5 / x

Multiply both sides above by x:
1.84*x / 1 = 42.5
1.84*x = 42.5
Divide both sides by 1.84:
x = 42.5 / 1.84

X = 23.10 mL

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3 years ago

Read 2 more answers

1.What percent of light passes through the sample if its absorbance A=2?

Dennis_Churaev [7]

1. The formula for absorbance is given as:

A = log (Io / I)

where A is absorbance, Io is initial intensity, and I is final light intensity

log (Io / I) = A

log (Io / I) = 2

Io / I = 100

Taking the reverse which is I / Io:

I / Io = 1 / 100

I / Io = 0.01

Therefore this means that only 0.01 fraction of light or 1% passes through the sample.

2. What is meant by transmittance values is actually the value of I / Io. So calculating for A:

at 10% transmittance = 0.10

A = log (Io / I)

A = log (1 / 0.10)

A = 1

at 90% transmittance = 0.90

A = log (Io / I)

A = log (1 / 0.90)

A = 0.046

So the absorbance should be from 0.046 to 1

3. at 10% transmittance = 0.10

A = log (Io / I)

A = log (1 / 0.10)

A = 1

3 0

3 years ago

How much salt (NaCl) is carried by a river flowing at 30.0 m3/s and containing 50.0 mg/L of salt

denis23 [38]

Answer:

129,600kg/day

Explanation:

The river is flowing at 30.0

1 m³ is equivalent to 1000L

flowrate of river = 30*1000 =30,000L/s

Convert L/s to litre per day by multiplying by 24*60*60

flowrate of river = 30,000 * 24*60*60 L/day

= 2,592,000,000L/day

if the river contains 50mg of salt in 1L of solution

lets find how many mg of salt Y is contained in 2,592,000,000L/day

by cross multiplying we have

Y= $\frac{2592000000*50}{1}$

Y= 129,600,000,000 mg/day

convert this value to kg/day by dividing by 1 million

Y= 129,600,000,000/1000000

Y= 129,600kg/day

3 0

3 years ago