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9 months ago

Please help, How could you tell a Ca(NO3)2 solution from a Zn(NO3)2 solution?

Chemistry

1 answer:

tamaranim1 [39]9 months ago

3 0

Answer:

you better give me brainliest

Explanation:

Zinc nitrate and calcium nitrate solution can be distinguished by reaction with ammonium hydroxide. Zinc forms a white gelatinous ppt. whereas there is no precipitation of calcium hydroxide even with excess of ammonium hydroxide

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100 POINTS ANSWER HURRRY

sleet_krkn [62]

Answer is A. The reason to create such is to conduct electricity, and metal conducts electricity.

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2 years ago

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Which is composed of alkynes?

telo118 [61]

The correct answer is letter (A) Acetylene. Acetylene is the most simplest form of alkyne and at the same time a hydrocarbon. It is unsaturated because of the presence of only two carbon atoms that are bonded together in a triple bond. In its pure, it is unstable and thus, it is usually held and handled as a solution.

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Draw the ring chain of isomer of propene ?

aksik [14]

Explanation:

Molecular formula for Propene = C3H6

The isomer of propene is cyclopropane.

(Draw a triangle to show that it is cyclopropane)

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2 years ago

Calculate the volume in milliliters of a 0.420mol / L barlum chlorate solution that contains 25.0 g of barium chlorate (Ba(ClO 3

Pachacha [2.7K]

Answer: The volume of barium chlorate is 195.65 mL

Explanation:

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

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

Given mass of barium chlorate = 25.0 g

Molar mass of barium chlorate = 304.23 g/mol

Molarity of solution = 0.420 mol/L

Volume of solution = ?

Putting values in above equation, we get:

$0.420mol/L=\frac{25.0\times 1000}{304.23\times V}\\\\V=\frac{25.0\times 1000}{304.23\times 0.420}=195.65mL$

Hence, the volume of barium chlorate is 195.65 mL

6 0

3 years ago

A sample of xenon gas occupies a volume of 6.80 L at 52.0°C and 1.05 atm. If it is desired to increase the volume of the gas sam

Pavlova-9 [17]

Answer:

207.03°C

Explanation:

The following data were obtained from the question:

V1 (initial volume) = 6.80 L

T1 (initial temperature) = 52.0°C = 52 + 273 = 325K

P1 (initial pressure) = 1.05 atm

V2 (final volume) = 7.87 L

P2 (final pressure) = 1.34 atm

T2(final temperature) =?

Using the general gas equation P1V1/T1 = P2V2/T2, the final temperature of the gas sample can be obtained as follow:

P1V1/T1 = P2V2/T2

1.05 x 6.8/325 = 1.34 x 7.87/T2

Cross multiply to express in linear form as shown below:

1.05 x 6.8 x T2 = 325 x 1.34 x 7.87

Divide both side by 1.05 x 6.8

T2 = (325 x 1.34 x 7.87) /(1.05 x 6.8)

T2 = 480.03K

Now, let us convert 480.03K to a number in celsius scale. This is illustrated below:

°C = K - 273

°C = 480.03 - 273

°C = 207.03°C

Therefore, the final temperature of the gas will be 207.03°C

5 0

3 years ago