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Anarel [89]
3 years ago
13

Brianna wanted to compare the densities of two different solids of the same size 6cm3 (same volume) to see which one was more de

nse. Solid A had a mass of 2 grams and Solid B had a mass of 0.5 grams. (Density = Mass/volume)
this has to be done in CER form
Chemistry
2 answers:
Lelu [443]3 years ago
7 0

Answer:

Density of Solid A is greater than B

Explanation:

<u>Given:</u>

Mass of Solid A = 2 g

Mass of solid B = 0.5 g

Volume of both solids = 6 cm3

<u>To determine:</u>

Densities of A and B

<u>Calculation:</u>

Density of a substance is the mass occupied by it in unit volume

Density = \frac{Mass}{Volume}

For Solid A:

Density = \frac{2\ g}{6\ cm3} =0.33\ g/cm3

For Solid B :

Density = \frac{0.5\ g}{6\ cm3} =0.083\ g/cm3

Therefore solid A is more dense than solid B

OLga [1]3 years ago
5 0
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Answer:

1)  Ethanol

Explanation:

If we will have <u>interactions</u> we will need more <u>energy</u> to break them in order to go from liquid to gas. If we need more <u>energy</u>, therefore, the <u>temperature will be higher</u>.

In this case, we can discard the <u>propanone</u> because this molecule don't have the ability to form <u>hydrogen bonds</u>. (Let's remember that to have hydrogen bonds we need to have a hydrogen bond to a <u>heteroatom</u>, O, N, P or S).

Then we have to analyze the hydrogen bonds formed in the other molecules. For ethanol, we will have only <u>1 hydrogen bond</u>. For water and ethanoic acid, we will have <u>2 hydrogen bonds</u>, therefore, we can discard the ethanol.  

For ethanoic acid, we have 2 <u>intramolecular hydrogen bonds</u>. For water we have 2 <u>intermolecular hydrogen bonds</u>, therefore, the strongest interaction will be in the <u>ethanoic acid</u>.

The<u> closer boiling point</u> to the 75ºC is the <u>ethanol</u> (boiling point of 78.8 ºC) therefore these molecules would have <u>enough energy</u> to <u>break</u> the hydrogen bonds and to past from<u> liquid to gas</u>.

4 0
2 years ago
According to the Brinsted -Lowry. what is an acid ? what is a base ? ​
sasho [114]

Answer:

According to the Brønsted definition, an acid is a substance capable of donating a proton, and a base is a substance capable of accepting a proton. ... The species giving up the proton is HCl, an acid. The species accepting the proton is water, the base. The species Cl- is the conjugate base of HCl.

8 0
3 years ago
4. When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0˚C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25.0˚C in a calorimeter,
myrzilka [38]

Answer:

The final temperature of the mixture is 28.11 °C

Explanation:

Step 1: Data given

Volume of 1.00 M Ba(NO3)2 = 1.00 L

Temperature = 25.0 °C

Volume of 1.00 M Na2SO4 = 1.00 L

enthalpy change is – 26 kJ per mol BaSO4

The specific heat of water is 4.18 J/g ·˚C

the density of water is 1.00 g/mL

Step 2: The balanced equation

Ba(NO3)2(aq) + Na2SO4(aq) → 2NaNO3(aq) + BaSO4(s)

Step 3: Calculate the total volume

Total volume = 1.00 L + 1.00 L = 2.00 L = 2000 mL

Step 4: Calculate mass

Mass = volume * density

Mass = 2000 mL * 1g/mL

Mass = 2000 grams

Step 5: Calculate moles BaSO4 formed

For 1 mol Ba(NO3)2 we need 1 mol Na2SO4 to produce 1 mol BaSO4

There is no limiting reactant, both Ba(NO3)2 and Na2SO4 will be completely be consumed (1 mol). We'll have 1.0 mol of BaSO4 produced.

Step 6: Calculate Q

Q = - ΔH

ΔH is negative so the reaction is exothermic, what means the temperature increases

Q is always positive, so Q = 26kJ = 26000 J

Step 6: Calculate the heat transfer

Q= m*c*ΔT

⇒with Q = the heat transfer = TO BE DETERMINED

⇒with m =the mass of the solution = 2000 grams

⇒with c= the specific heat of the solution = 4.18 J/g°C

⇒with ΔT = the change of temperature = T2 - T1 = T2 - 25.0

26000 = 2000 * 4.18 * (T2 - 25.0 °C)

3.11 = T2 - 25.0 °C

T2 = 25.0 + 3.11 °C

T2 = 28.11 °C

The final temperature of the mixture is 28.11 °C

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3 years ago
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NeTakaya

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2 years ago
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Maurinko [17]

Answer:

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Explanation:

From the ionic equation

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Only sodium is the spectator ion, so it cancels out, since C6H5COOH and HCN do not ionize completely they are left undissociated

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