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Crazy boy [7]
2 years ago
7

Solids retain their shape and size, if kept constant temperature and pressure. At a constant temperature and pressure, liquids r

etain their:
a. volume
b. shape
c. size
d. shape and volume
Chemistry
2 answers:
snow_tiger [21]2 years ago
8 0
Simply volume, as shape and size depend on the host container

Gnoma [55]2 years ago
5 0
At a constant temperature and pressure, liquids retain their volume
You might be interested in
A gas sample occupies 3.50 liters of volume at 20.°c. what volume will this gas occupy at 100.°c (reported to three significant
VLD [36.1K]

Explanation:

According to Charle's law, at constant pressure the volume of an ideal gas is directly proportional to the temperature.

That is,             Volume \propto Temperature

Hence, it is given that V_{1} is 3.50 liters, T_{1} is 20 degree celsius, and T_{2} is 100 degree celsius.

Therefore, calculate V_{2} as follows.

                           \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

                           \frac{3.50 liter}{20^{o}C} = \frac{V_{2}}{100^{o}C}

                                V_{2} = 17.5 liter

Thus, we can conclude that volume of gas required at 100 degree celsius is 17.5 liter.

6 0
2 years ago
Read 2 more answers
Draw a sodium formate molecule. The structure has been supplied here for you to copy. To add formal charges, click the button be
Karo-lina-s [1.5K]

The Molecule of Sodium Formate along with Formal Charges (in blue) and lone pair electrons (in red) is attached below.

Sodium Formate is an ionic compound made up of a positive part (Sodium Ion) and a polyatomic anion (Formate).

Nomenclature:

                       In ionic compounds the positive part is named first. As sodium ion is the positive part hence, it is named first followed by the negative part i.e. formate.

Name of Formate:

                             Formate ion has been derived from formic acid ( the simplest carboxylic acid). When carboxylic acids looses the acidic proton of -COOH, they are converted into Carboxylate ions.

E.g.

                    HCOOH (formic acid)    →     HCOO⁻ (formate)  +  H⁺

                H₃CCOOH (acetic acid)     →      H₃CCOO⁻ (acetate)  +  H⁺

Formal Charges:

                           Formal charges are calculated using following formula,

          F.C  =  [# of Valence e⁻] - [e⁻ in lone pairs + 1/2 # of bonding electrons]

For Oxygen:

                    F.C  =  [6] - [6 + 2/2]

                    F.C  =  [6] - [6 + 1]

                    F.C  =  6 - 7

                    F.C  =  -1

For Sodium:

                    F.C  =  [1] - [0 + 0/2]

                    F.C  =  [1] - [0]

                    F.C  =  1 - 0

                    F.C  =  +1

5 0
2 years ago
Complete combustion of a 0.600-g sample of a compound in a bomb calorimeter releases 24.0 kJ of heat. The bomb calorimeter has a
JulijaS [17]

Answer : The correct option is, 30.9^oC

Explanation :

Formula used :

q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

where,

q = heat released = 24 KJ

m = mass of bomb calorimeter = 1.30 Kg

c = specific heat = 3.41J/g^oC

T_{final} = final temperature = ?

T_{initial} = initial temperature = 25.5^oC

Now put all the given values in the above formula, we get  the final temperature of the calorimeter.

q=m\times c\times (T_{final}-T_{initial})

24KJ=1.30Kg\times 3.41J/g^oC\times (T_{final}-25.5)^oC

T_{final}=30.9^oC

Therefore, the final temperature of the calorimeter is, 30.9^oC

5 0
2 years ago
Read 2 more answers
A quantity of 85.0 mL of 0.900 M HCl is mixed with 85.0 mL of 0.900 M KOH in a constantpressure calorimeter that has a heat capa
bogdanovich [222]

Explanation:

The given data is as follows.

         V_{1} = 85.0 ml,        M_{1} = 0.9 M

         V_{2} = 85.0 ml,        M_{1} = 0.9 M

Hence, number of moles of HCl and KOH will be the same because both the solutions have same volume and molarity.

So,     No. of moles = Molarity × Volume

                                = 0.9 M \times 0.085 L        (as 1 L = 1000 ml so, 85 ml = 0.085 L)

                                = 0.076 mol

As 1 mole gives 56.2 kJ/mol of heat of neutralization. Hence, calculate the heat of neutralization given by 0.076 moles as follows.

              56.2 kJ/mol \times 0.076 mol

                    = 4.271 kJ

or,                 = 4271 J     (as 1 kJ = 1000 J)

Therefore,    heat released = - heat of gained by calorimeter

Since, it is given that density of the solution is similar to the density of water which is 1 g/ml.

Hence,     mass of HCl = density × Volume of HCl

                                      = 1.00 g/ml × 85.0 ml

                                       = 85 g

Similarly,    mass of KOH = = density × Volume of HCl

                                      = 1.00 g/ml × 85.0 ml

                                       = 85 g

Hence, total mass of the solution = 85 g + 85 g

                                                        = 170 g

Also,                   q = mC \Delta T

                     4271 J = 170 g \times 325 J/^{o}C \times (T_{f} - 18.24)^{o}C    

                     0.0773 = T_{f} - 18.24

                    T_{f} = 18.317^{o}C  

Thus, we can conclude that final temperature of the mixed solution is 18.317^{o}C.

6 0
2 years ago
For the reaction n2(g) + 2h2(g) â n2h4(l), if the percent yield for this reaction is 77.5%, what is the actual mass of hydrazine
Rudiy27

First calculate the moles of N2 and H2 reacted.

moles N2 = 27.7 g / (28 g/mol) = 0.9893 mol

moles H2 = 4.45 g / (2 g/mol) = 2.225 mol

 

We can see that N2 is the limiting reactant, therefore we base our calculation from that.

Calculating for mass of N2H4 formed:

mass N2H4 = 0.9893 mol N2 * (1 mole N2H4 / 1 mole N2) * 32 g / mol * 0.775

<span>mass N2H4 = 24.53 grams</span>

7 0
2 years ago
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