All categories

3 years ago

A sample of helium gas is heated from 20.0°C to 40.0°C. This

Chemistry

1 answer:

Montano1993 [528]3 years ago

8 0

Answer:

C. 548mL

Explanation:

Attached is an image of the explanation. Hope this helps!

You might be interested in

Three different samples were weighed using a different type of balance for each sample. The three were found to have masses of 0

Lena [83]

Answer:

6.096799125kg

Explanation:

According to the question, three different samples weighed using different types of balance had masses: 0.6160959 kg, 3.225 mg, and 5480.7 g.

Based on observation, the mass units in the three measurements are different but must be uniform in order to find the total mass. Hence, we need to convert to the standard unit (S.I unit of mass), which is kilograms (kg)

Since 1kg equals 1,000,000mg

Hence, 3.225mg will be 3.225/1000000

= 0.000003225kg

Also, 1kg equals 1000g

Hence, 5480.7g will be 5480.7/1000

= 5.4087kg

Hence, the total mass of the three samples (now in the same unit) are:

5.4807kg + 0.000003225kg + 0.6160959 kg

= 6.096799125kg

3 0

3 years ago

Dmitri mendeleev contribution to the periodic table.

Pepsi [2]

In 1871, a Russian Chemist, Dimitri Mendeleev, gave a useful scheme for classification of elements. He presented the first regular periodic table in which elements of similar chemical properties were arranged in eight vertical columns called groups. The horizontal rows of table were called periods. He arranged elements in ascending order of their atomic masses and found that elements having similar chemical properties appeared at regular intervals. This observation was called Periodic Law.

8 0

2 years ago

How many grams of precipitate will be formed when 20.5 mL of 0.800 M

Anton [14]

Answer:

There will be formed 1.84 grams of precipitate (NaNO3)

Explanation:

Step 1: The balanced equation

CO(NO3)2 (aq) + 2 NaOH (aq) → CO(OH)2 (s) + 2 NaNO3 (aq)

Step 2: Data given

Volume of 0.800 M CO(NO3)2 = 20.5 mL = 0.0205 L

Volume of 0.800 M NaOH = 27.0 mL = 0.027 L

Molar mass of NaNO3 = 84.99 g/mol

Step 3: Calculate moles of CO(NO3)2

Moles CO(NO3)2 = Molarity * volume

Moles CO(NO3)2 = 0.800 M * 0.0205

Moles CO(NO3)2 = 0.0164 moles

Step 4: Calculate moles NaOH

moles of NaOH = 0.800 M * 0.027 L

moles NaOH = 0.0216 moles

Step 5: Calculate limiting reactant

For 1 mole CO(NO3)2 consumed, we need 2 moles of NaOH to produce 1 mole of CO(OH)2 and 2 moles of NaNO3

NaOH is the limiting reactant. It will completely be consumed.

CO(NO3)2 is in excess. There willbe 0.0216 / 2 = 0.0108 moles of CO(NO3)2 consumed. There will remain 0.0164 - 0.0108 = 0.0056 moles of CO(OH)2

Step 6: Calculate moles of NaNO3

For 2 moles of NaOH consumed, we have 2 moles of NaNO3

For 0.0216 moles of NaOH, we have 0.0216 moles of NaNO3

Step 7: Calculate mass of NaNO3

mass of NaNO3 = moles of NaNO3 * Molar mass of NaNO3

mass of NaNO3 = 0.0216 moles * 84.99 g/mol = 1.84 grams

There will be formed 1.84 grams of precipitate (NaNO3)

5 0

3 years ago

In a laboratory experiment, John uses a mesh to separate soil particles from water. Which technique of separation is he using?

Aneli [31]

Salutations!

In a laboratory experiment, John uses a mesh to separate soil particles from water. Which technique of separation is he using?

The technique that John is using is the filtration technique. Filtration is a technique to separate the solid which is insoluble from the liquid. For instance: Sand and water, sand is insoluble, thus it stays in the filter paper, while the water proceeds through the filter paper.

Hope I helped :D

3 0

3 years ago

How many grams of potassium nitrate (KNO3) would form if 2.25 liters of a 1.50 molar lead nitrate Pb(NO3)2 solution reacts with

fgiga [73]

I know that the answer is 639 g

4 0

3 years ago