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musickatia [10]
3 years ago
6

The combination reagents that can be used to distinguish between ethane, ethene and ethyne are?​

Chemistry
1 answer:
mafiozo [28]3 years ago
5 0

Answer:

Two reagents that can be used to distinguish between ethane and ethyne are Tollens' reagent and ammoniacal CuCl solution.

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What is the fuel used in a fuel-cell automobile
Katena32 [7]
Hydrogen gas

Explanation

6 0
3 years ago
A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
Ede4ka [16]

<u>Answer:</u> The mass of glucose in final solution is 1.085 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}      ......(1)

Given mass of glucose = 15.5 g

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

\text{Molarity of glucose solution}=\frac{15.5\times 1000}{180.2\times 100}\\\\\text{Molarity of glucose solution}=0.860M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated glucose solution

M_2\text{ and }V_2 are the molarity and volume of diluted glucose solution

We are given:

M_1=0.860M\\V_1=35.0mL\\M_2=?M\\V_2=0.500L=500mL

Putting values in above equation, we get:

0.860\times 35.0=M_2\times 500\\\\M_2=\frac{0.860\times 35.0}{500}=0.0602M

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of glucose solution = 0.0602 M

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

0.0602=\frac{\text{Mass of glucose solution}\times 1000}{180.2\times 100}\\\\\text{Mass of glucose solution}=\frac{0.0602\times 180.2\times 100}{1000}=1.085g

Hence, the mass of glucose in final solution is 1.085 grams

4 0
3 years ago
What is the scientific method?
OLEGan [10]
D. A set of steps that help scientists gather knowledge
6 0
3 years ago
How much energy is released by the decay of 3 grams of 230Th in the following reaction 230 Th - 226Ra + "He (230 Th = 229.9837 g
Serhud [2]

<u>Answer:</u> The energy released for the decay of 3 grams of 230-Thorium is 2.728\times 10^{-15}J

<u>Explanation:</u>

First we have to calculate the mass defect (\Delta m).

The equation for the alpha decay of thorium nucleus follows:

_{90}^{230}\textrm{Th}\rightarrow _{88}^{226}\textrm{Ra}+_2^{4}\textrm{He}

To calculate the mass defect, we use the equation:

Mass defect = Sum of mass of product - Sum of mass of reactant

\Delta m=(m_{Ra}+m_{He})-(m_{Th})

\Delta m=(225.9771+4.008)-(229.9837)=1.4\times 10^{-3}amu=2.324\times 10^{-30}kg

(Conversion factor: 1amu=1.66\times 10^{-27}kg )

To calculate the energy released, we use Einstein equation, which is:

E=\Delta mc^2

E=(2.324\times 10^{-30}kg)\times (3\times 10^8m/s)^2

E=2.0916\times 10^{-13}J

The energy released for 230 grams of decay of thorium is 2.0916\times 10^{-13}J

We need to calculate the energy released for the decay of 3 grams of thorium. By applying unitary method, we get:

As, 230 grams of Th release energy of = 2.0916\times 10^{-13}J

Then, 3 grams of Th will release energy of = \frac{2.0916\times 10^{-13}}{230}\times 3=2.728\times 10^{-15}J

Hence, the energy released for the decay of 3 grams of 230-Thorium is 2.728\times 10^{-15}J

5 0
2 years ago
In the covalent compound C3H8 the Greek prefix used to represent the cation is?
krok68 [10]

Answer:

pro

Explanation:

c3h8 is propane

3 carbons makes it PROpane

the ANE come from all single bonds

4 0
3 years ago
Read 2 more answers
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