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3 years ago

How is aklkene cornveted to alkane?

1 answer:

3 0

H2 adds across double (or triple bonds) in the presence of a suitable catalyst toconvert an alkene (or alkyne) to the corresponding alkane.

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The element that has a valence configuration of 5s2 is Ba Ca Be Sr Mg

marysya [2.9K]

Answer: Sr

Explanation: Strontium electron configuration is [Kr] 5s2. It belongs to group 2 and period 5 of the periodic table.

5 0

2 years ago

A chemist reacted 17.25 grams of sodium metal with an excess amount of chlorine gas. The chemical reaction that occurred is show

Afina-wow [57]

Answer: The actual yield of the product is 38.57 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

We are given:

Given mass of Na = 17.25 g

Molar mass of Na = 23 g/mol

Putting values in equation 1, we get:

$\text{Moles of Na}=\frac{17.25g}{23g/mol}=0.75mol$

For the given chemical reaction:

$2Na+Cl_2\rightarrow 2NaCl$

By stoichiometry of the reaction:

2 moles of Na produces 2 moles of NaCl

So, 0.75 moles of Na will produce = $\frac{2}{2}\times 0.75=0.75mol$ of NaCl

We know, molar mass of $NaCl$ = 58.44 g/mol

Putting values in above equation, we get:

$\text{Mass of NaCl}=(0.75mol\times 58.44g/mol)=43.83g$

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100$ ......(1)

Given values:

Percent yield of the product = 88 %

Theoretical value of the product = 43.83 g

Plugging values in equation 1:

$88\% =\frac{\text{Actual yield}}{43.83g}\times 100\\\\\text{Actual yield}=\frac{88\times 43.83}{100}=38.57g$

Hence, the actual yield of the product is 38.57 g

3 0

3 years ago

a solution with a transmittance of 0.44 is analyzed in a spectrophotometer with 6% stray light. calculate the absorbance reporte

IgorLugansk [536]

The absorbance reported by the defective instrument was 0.3933.

Absorbance A = - log₁₀ T

Tm = transmittance measured by spectrophotometer

Tm = 0.44

Absorbance reported in this equipment = -log₁₀ (0.44) = 0.35654

True absorbance can be calculated by true transmittance, Tm = T+S(α-T)

S = fraction of stray light = 6%= 6/100 = 0.06

α= 1, ideal case

T = true transmittance of the sample

Tm = T+S(α-T)

now, T= Tm-S/ 1-S = 0.44-0.06/ 1-0.06 = 0.404233

therefore, actual reading measured is A = -log₁₀ T = -log₁₀ (0.404233)

i.e; 0.3933

To know more about transmittance click here:

brainly.com/question/17088180

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3 0

2 years ago

Which of the following is an example of a buffer?

Aleonysh [2.5K]

It's Hemoglobin- i just took test

3 0

4 years ago

1.45 moles of helium gas occupies a volume 18.5 L at a pressure of .85 atm. What is the temperature of the gas in K and °C?

Anon25 [30]

Answer: The temperature of the gas is 132.25K and -140.75°C

Explanation:

Given that:

Volume of helium gas (V) = 18.5L

Temperature of helium gas (T) = ?

Pressure (P) = 0.85 atm

Number of moles (n) = 1.45 moles

Molar gas constant (R) is a constant with a value of 0.082 atm L K-1 mol-1

Then, apply the formula for ideal gas equation

pV = nRT

0.85atm x 18.5L = 1.45 moles x 0.082 atm L K-1 mol-1 x T

15.725 atm•L = 0.1189 atm•L•K-1 x T

T = 15.725 atm•L / 0.1189 atm•L•K-1

T = 132.25K

Given that temperature in Kelvin is 132.25, convert it to Celsius by subtracting 273

So, T(in Celsius) = 132.25K - 273

= -140.75°C

Thus, the temperature of the gas is 132.25K and -140.75°C

5 0

4 years ago