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Vlada [557]
2 years ago
14

To save time you can approximate the initial volume of water to ±1 mL and the initial mass of the solid to ±1 g. For example, if

you are asked to add 23 mL of water, add between 22 mL and 24 mL. Which metals in each of the following sets will have equal density?
1. 20.2 g gold placed in 21.6 mL of water and 12.0 g copper placed in 21.6 mL of water.
2. 20.2 g silver placed in 21.6 mL of water and 12.0 g silver placed in 21.6 mL of water.
3. 15.2 g copper placed in 21.6 mL of water and 50.0 g copper placed in 23.4 mL of water.
4. 15.4 g gold placed in 20.0 mL of water and 15.7 g silver placed in 20.0 mL of water.
5. 20.2 g silver placed in 21.6 mL of water and 20.2 g copper placed in 21.6 mL of water.
6. 11.2 g gold placed in 21.6 mL of water and 14.9 g gold placed in 23.4 mL of water.
Chemistry
1 answer:
N76 [4]2 years ago
7 0

Answer:

The correct answers are option 2, 3 and 6.

Explanation:

Density is defined as mas of substance present in an unit volume of the substance.

Density=\frac{Mass}{Volume}

Density of same substance with different masses and volume remains the same that is it is an intensive property.

2. 20.2 g silver placed in 21.6 mL of water and 12.0 g silver placed in 21.6 mL of water.

3. 15.2 g copper placed in 21.6 mL of water and 50.0 g copper placed in 23.4 mL of water.

6. 11.2 g gold placed in 21.6 mL of water and 14.9 g gold placed in 23.4 mL of water.

Since the metal kept in both the cases are same.And metal has fix value of density , so from the given options the the option with sets of same of  metals has equal densities.

You might be interested in
A birthday balloon had a volume of 14.1 L when the gas inside was at a temperature of 13.9 °C. Assuming no gas escapes, what is
bixtya [17]

Answer:

14.5L

Explanation:

The following data were obtained from the question:

V1 = 14.1L

T1 = 13.9°C = 13.9 + 273 = 286.9K

T2 = 22°C = 22 + 273 = 295K

V2 =?

Using charles' law: V1/T1 = V2 /T2, we can obtain the new volume as follows:

14.1/286.9 = V2 /295

Cross multiply to express in linear form

286.9 x V2 = 14.1 x 295

Divide both side by 286.9

V2 = (14.1 x 295) / 286.9

V2 = 14.5L

Therefore, the new volume = 14.5L

8 0
3 years ago
Hydrogen reacts with chlorine to form hydrogen chloride (HCl (g), Delta.Hf = –92.3 kJ/mol) according to the reaction below. Uppe
erik [133]

Answer:

The enthalpy of the reaction is –184.6 kJ, and the reaction is exothermic.

Explanation:

7 0
2 years ago
Read 2 more answers
Explain the effects of nh3 and hcl on the cuso4 solution in terms of le chatelier's principle
Fittoniya [83]

The Principle of Le Chatelier states that if a system in equilibrium is subjected to a disturbance, the system will react in such a way that it will diminish the effect of that disturbance. Thus, when the concentration of one of the substances in an equilibrium system is changed, the equilibrium varies in such a way that it can compensate for this change.

For example, if the concentration of one of the reactants is increased, the equilibrium shifts to the right or to the side of the products. Also, if you add more reagents, the reaction will move even more to the right until the balance is re-established again, increasing the quantity of products.

In this way, adding HCl to a solution of CuSO4 will produce the following reaction:

CuSO4 (aq) + 2HCl (aq) ⇔ CuCl2 (aq) + H2SO4 (aq)

Initially the solution of CuSO4 in water will be blue, but when adding HCl the solution will change color to green, since the aqueous solutions of CuCl2 are green. By adding more HCl this color will intensify as the balance shifts to the right, producing more CuCl2 and H2SO4.

On the other hand, adding NH3 to a solution of CuSO4 will produce the following reaction

CuSO4 (aq) + 4NH3 (aq) ⇔ [Cu(NH3)4] SO4 (s)

Thus, by adding NH3 to the CuSO4 solution we will observe the formation of a precipitate corresponding to [Cu(NH3) 4] SO4. <u>When adding more NH3, the formation of more precipitate will be observed as the equilibrium moves to the right, producing a greater quantity of [Cu (NH3) 4] SO4.</u>

6 0
2 years ago
The normal boiling point of acetic acid is 118.1°C. If a sample of the acetic acid is at 125.2°C, predict the signs of ΔH, ΔS, a
Nostrana [21]

The question is incomplete, the complete question is;

The normal boiling point of acetic acid is 118.1°C. If a sample of the acetic acid is at 125.2°C, predict the signs of ∆H, ∆S, and ∆G for the boiling process at this temperature.

A. ∆H > 0, ∆S > 0, ∆G < 0

B. ∆H > 0, ∆S > 0, ∆G > 0

C. ∆H > 0, ∆S < 0, ∆G < 0

D. ∆H < 0, ∆S > 0, ∆G > 0

E. ∆H < 0, ∆S < 0, ∆G > 0

Answer:

A. ∆H > 0, ∆S > 0, ∆G < 0

Explanation:

During boiling, a liquid is converted to vapour. This is a phase change for which heat is absorbed because energy must be taken in to break the intermolecular bonds in the liquid before it can be converted to a gas. Hence ∆H>0

Secondly, a phase change from liquid to gas leads to an increase in entropy hence ∆S>0.

Thirdly, the process is spontaneous. For every spontaneous process ∆G<0

3 0
2 years ago
A. Calculations for the Determination of Ammonium Chloride The data from the data entry portion of the report has been copied in
Vladimir79 [104]

Answer:

A

Explanation:

Considering question A

Mass of  original sample is m_o = 0.945 \ g

Mass of   NH4Cl is  m_n = 0.116 \ g

Percent of  NH4Cl is k   =  12.275 \%

B

Mass of  NaCl  m_k  =  0.359 \ g

C

Mass  of  SiO2   m_e = 0.46

D

 Mass of original sample m_o = 0.945 \ g

  Differences in these weights (g) (use the absolute value of the difference)

recovery of matter   G  =   0.01 \ g

The  correct option is C

From the question we are told that

The mass of evaporating dish on #1 is  m_1 =  38.646 \ g

 The mass of evaporating dish and original sample   m_2 =  39 591 \ g

  The mass of evaporating dish after subliming NH_4Cl is m_3 =  39.4750 \  g

Generally the mass of the original sample is  mathematically represented as

        m_o =  m_2 - m_1

=>     m_o =  39 591 -  38.646

=>     m_o = 0.945 \ g

Generally the mass of NH_4Cl is mathematically represented as

        m_n = m_2 - m_3

=>      m_n = 39 591 - 39.4750

=>      m_n = 0.116 \ g

The  Percent  NaH_4 Cl (g)

        k   =  \frac{ m_n}{m_o} *100

=>     k   =  \frac{0.116 }{0.945} *100

=>      k   =  12.275 \%

Considering question B

The  mass of evaporating dish #2 is  m_g  =  38700\  g

The  mass of  watch glass is   m_a  =  28 299 \  g

The mass of evaporating dish #2, watch glass and NaCl  m_b  =  67,355 \  g

Generally the mass of NaCl is  

       m_k  =  m_b -[m_g + m_a]

=>     m_k  =  67,355  -[38700 + 28 299]

=>      m_k  =  0.359 \ g

Considering question C

 The mass of evaporating dish is   m_p= 38.645

 The mass of evaporating dish and SiO2     m_s  = 39.105 \ g

Generally  the mass of  SiO2  is  mathematically represented as

        m_e = 39.105 - 38.645

=>      m_e = 0.46

Considering  D

The  mass of the original  sample  is  m_o  =  0.945 \  g

Generally the experimental  mass recovered (NH_4Cl,NaCl, SiO2 ) is mathematically evaluated as

     M =0.116 + 0.46 + 0.359

    M  =  0.935 \ g

Generally the differences in these weights (g) of recovery of matter is mathematically represented as.

     G  =0.945- 0.935

=>   G  =   0.01 \ g

While drying the NaCl, the liquid boiled and some splattered out of the evaporating dish, causing the recovered mass to be lower.

     

6 0
2 years ago
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