answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
gayaneshka [121]
2 years ago
7

If 10.0 liters of h2(g) at STP is heated to a temperature of 546 K, pressure remaining constant, the new volume of the gas will

be
Chemistry
1 answer:
Effectus [21]2 years ago
3 0

Answer:

20L

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 10L

Initial Temperature (T1) = stp = 273K

Final temperature (T2) = 546K

Final volume (V2) =..?

The new volume of the gas can be obtained by using the Charles' law equation as shown below:

V1/T1 = V2/T2

10/273 = V2/546

Cross multiply to express in linear form

273 x V2 = 10 x 546

Divide both side by 273

V2 = (10 x 546) / 273

V2 = 20L

Therefore, the new volume of the gas is 20L.

You might be interested in
A compound with the molecular formula C10H10O4 produces a 1H NMR spectrum that exhibits only two signals, both singlets. One sig
Kamila [148]

Answer:

The attached figure shows the structure of dimethyl terephthalate.

Explanation:

Dimethyl terephthalate is a compound whose formula is C6H4 (COOCH3) 2. It is a diester produced from terephthalic acid and methanol. It is characterized by being a white solid. Another method for the preparation is from p-xylene and methanol, which is characterized by having an oxidation and an esterification.

8 0
2 years ago
The lab procedure involves several factors, listed below. Some were variable and some were constant. Label each factor below V f
Mariulka [41]

Answer :

V - mass of the water in the calorimeter

V - mass of the metal

V - change in temperature of the water

V - change in temperature of the metal

C - volume of water in calorimeter

C - calorimeter pressure

C - specific heat of water

Explanation :

Variables : It is a factor that changes during the experiment or calculation.

Constant : It is a factor that does not change during the experiment or calculation.

In a calorimeter, the heat absorbed is equal to the heat released.

Q_{absorbed}=Q_{released}

As we know that,  

Q=m\times c\times \Delta T

m_1\times c_1\times \Delta T_1=-[m_2\times c_2\times \Delta T_2]

where,

m_1 = mass of water in calorimeter

m_2 = mass of metal

\Delta T_1 = change in temperature of the water

\Delta T_2 = change in temperature of the metal

c_1 = specific heat of water

c_2 = specific heat of metal

From this, we conclude that the value of specific heat of water is constant while the other are variables.

The volume of water in calorimeter, calorimeter pressure is also constant.

8 0
1 year ago
Read 2 more answers
A 25.0-g sample of ice at -6.5oC is removed from the freezer and allowed to warm until it melts. Given the data below, select al
alukav5142 [94]

Answer:

B, D

Explanation:

The strategy here is to realize that the ice will be taken from -6.5 ºC to OºC where it will melt.

Lets call q₁ the heat required to bring the ice to 0ºc, q₂ the heat required to bring the phase change from solid to liquid.

q₁ is calculated from the expression

q₁ = s x m x ΔT where m is the mass, s the specific heat of ice ( 2.09 J/gºC ) and ΔT  the change in temperature.

q₂ the fusion enthalpy change   is calculated from the expression:

q₂ = C x ΔT

where C is the specific heat for the phase change , in this case named AH  given in kJ/mol.

We are given all the data needed to calculate q₁, q₂ and qtotal ( q₁ + q₂ )

q₁ = 25.0 g x ( 2.09 J/gºC) x ( 0 - ( -6.5 ºC ) )

q₁ = 339.6 J = 0.339 kJ

q₂ = (25 g/ 18 g/mol) x 6.02 kJ/mol = 1.39 x 6.02 kJ = 8.36 kJ

qtotal = 0.339 kJ + 8.36 kJ = 8.70 kJ

with these calculations, we can now proceed to answer the question:

(a) False AH is theheat capacity for the melting.

(b) True as we determined above

(c) False we only have one phase change, from solid (ice) to liquid

(d) True as calculated above

(e) False as determined in our calculations

7 0
1 year ago
Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points. NH4l, CoBr3, Na2
seraphim [82]
Answer:

NaI > Na2SO4 > Co Br3

meaning that NaI has the highest freezing point, and Co Br3 has the lowest freezing point.

Explanation:

The freezing point depression is a colligative property.

That means that it depends on the number of solute particles dissolved.

The formula to calculate the freezing point depression of a solution of a non volatile solute is:

ΔTf = i * Kf * m

Where kf is a constant, m is the molality and i is the van't Hoff factor.

Molality, which is number of moles per kg of solvent, counts for the number of moles dissolved and the van't Hoff factor multipllies according for molecules that dissociate.

The higher the number of molecules that dissociate, the higher the van't Hoff, the greater the freezing point depression and the lower the freezing point.

As the question states that you assume equal concentrations (molality) and complete dissociation you just must find the number of ions generated by each solute, in this way:

NH4 I → NH4(+) + I(-) => 2 ions

Co Br3 → Co(+) + 3 Br(-) => 4 ions

Na2SO4 → 2Na(+) + SO4(2-) => 3 ions.

So, Co Br3 is the solute that generate more particles and that solution will exhibit the lowest freezing point among the options given, Na2SO4 is next and the NaI is the third. Ordering the freezing point from higher to lower the rank is NaI > Na2SO4 > CoBr3, which is the answer given.
4 0
1 year ago
Under standard conditions, a given reaction is endergonic (i.e., ΔG >0). Which of the following can render this reaction favo
sleet_krkn [62]

Answer:

Maintaining a high starting-material concentration can render this reaction favorable.

Explanation:

A reaction is <em>favorable</em> when <em>ΔG < 0</em> (<em>exergonic</em>). ΔG depends on the temperature and on the reaction of reactants and products as established in the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature

Q is the reaction quotient

To make ΔG < 0 when ΔG° > 0 we need to make the term R.T.lnQ < 0. Since T is always positive we need lnQ to be negative, what happens when Q < 1. Q < 1 implies the concentration of reactants being greater than the concentration of products, that is, maintaining a high starting-material concentration will make Q < 1.

5 0
1 year ago
Other questions:
  • A sample contains 2.2 g of the radioisotope niobium-91 and 15.4 g of its daughter isotope, zirconium-91. how many half-lives hav
    12·2 answers
  • What products are formed from monochlorination of (2r)−2−bromobutane at c1 and c4? draw the products using skeletal structures?
    10·1 answer
  • A chemical reaction happens
    9·1 answer
  • For the following equilibrium system, which of the following changes will form more CaCO3? CO2(g) + Ca(OH)2(s) ⇌ CaCO3(s) + H2O(
    14·1 answer
  • Be sure to answer all parts. Consider the formation of ammonia in two experiments. (a) To a 1.00−L container at 727°C, 1.30 mol
    6·2 answers
  • Suppose two chemical reactions are linked together in a way that the O2 produced in the first reaction goes on to react complete
    12·1 answer
  • Treating (CH3)3C-Cl with a mixture of H2O and CH3OH at room temperature would yield: A) CH2=C(CH3)2 B) (CH3)3COH C) (CH3)3COCH3
    11·1 answer
  • how are pure carbon, oxygen and hydrogen different from the compound ethanol which contains all three of those elements?
    14·1 answer
  • A scientist works for a drug company and runs experiments on their product with results that favor the company's product. This i
    9·1 answer
  • How are seat belts related to inertia?
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!