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loris [4]
3 years ago
14

If 6.5 L of water is added to 4.5 L of a 2.8 M Ca(OH)2 solution, what is the molarity of the new solution?

Chemistry
1 answer:
Mila [183]3 years ago
6 0
Find moles of Ca(OH)2 first:

moles = 2.8 x 4.5 = 12.6

Now,
   After addition, new volume =  6.5 + 4.5 = 11.0

So, new molarity =  12.6 / 11 = 1.145
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What volume of so2 is produced at 325 k and 1.35 atm when 15.0 grams of hcl reacts with excess k2so3?
vova2212 [387]
The volume of SO2 produced at 325k   is calculated as  below

calculate  the moles of SO2 produced  which  is calculated as follows

write the  reacting equation
K2SO3 +2 HCl = 2KCl +H2O+ SO2

find the   moles  of  HCl  used
=mass/molar mass = 15g/ 36.5 g/mol =0.411 moles

by  use of mole ratio between  HCl to  SO2  which is  2:1 the moles of SO2 is therefore = 0.411 /2 =0.206  moles  of SO2

use the idea  gas  equation  to calculate the volume SO2
that is V=nRT/P  
where  n=0.206  moles
          R(gas constant) = 0.082 L.atm/ mol.k
         T=325 K
          P=1.35 atm

V=(0.206 moles x  0.082 L.atm/mol.k x325 k)/1.35 atm = 4.07 L of SO2

3 0
2 years ago
Write a balanced chemical equation for the standard formation reaction of liquid acetic acid hch3co2.
Afina-wow [57]

The balanced chemical equation for the standard formation reaction of liquid acetic acid is given as ,

2C(gr) +2H_{2} (g) +O_{2} (g)  → CH_{3} COOH(l)

The reaction that form the products from their elements in their standard state is called formation of reaction .The acetic acid consist C , H , and O , So, determine their standard state . Carbon is graphite at 25°C and 1 atm , whereas hydrogen and oxygen are diatomic gases . Hence , we start with unbalanced reaction.

C(gr) +H_{2} (g) +O_{2} (g) → CH_{3} COOH(l)

The balanced chemical equation for the standard formation reaction of liquid acetic acid as,

2C(gr) +2H_{2} (g) +O_{2} (g)  → CH_{3} COOH(l)

The combustion of liquid acetic acid is given as,

CH_{3} COOH(l) + 2O(g) → 2CO_{2}((g) +2H_{2} O(l)                    ΔH =-873

learn more about  balancing chemical equation

brainly.com/question/15052184

#SPJ4

5 0
2 years ago
Calculate the amount of heat needed to boil of benzene (), beginning from a temperature of . Be sure your answer has a unit symb
Pavlova-9 [17]

Answer:

Check explanation

Explanation:

From the question, the parameters given are 64.7g of benzene,C6H6; a starting temperature of 41.9°C and bringing it to 33.2°C.

Molar mass of benzene,C6H6= 78.11236 g/mol.

Things to know: heat capacity of benzene, C6H6= 1.63 J/g.K, the heat of fusion = 9.87 kj/mol.

STEP ONE(1): ENERGY USED IN MELTING BENZENE SOLID.

Using the formula below;

Energy used in melting the solid(in JOULES) = (mass of benzene/molar mass of benzene) × heat of Fusion.

=(64.7 g of C6H6/ 78.11236(g per mol) of C6H6) × 9.87 kJ per mol.

= 8.175 J.

= 0.008175 kJ.

STEP TWO (2): ENERGY OF HEATING THE LIQUID.

It can be calculated from the formula below;

Energy= heat capacity (J/g.K) × mass of benzene× (∆T).

= 1.63 J/g.K × 64.7 × (41.9-33.2).

= 917.5J.

= 0.9175 kJ.

Energy required to boil benzene= Energy required to melt the bezene + energy required for boiling.

= 0.008175+ 0.9175.

= 0.93kJ

Approximately, 1 kJ

8 0
2 years ago
Can anybody answer this question of chemistry?
maw [93]

Answer:

Answer:A

Answer:AExplanation:

Answer:AExplanation:Molar Mass of glucose = (6×12)+(1×12)+(16×6)= 180g/mol

= 180g/molNumber of moles of Glucose = Mass/Molar Mass

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778moles

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556moles

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles={(12×2)+(1×6)+16} × 55.5556

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles={(12×2)+(1×6)+16} × 55.5556= 46.5×55.5556

= 180g/molNumber of moles of Glucose = Mass/Molar Mass= 5000/180= 27.7778molesIn the balanced equation of fermentation, the ratio of glucose to ethanol is 2:1Therefore the number of moles of ethanol is 2×27.7778=55.5556molesMass of ethanol= Molar Mass of ethanol × Number of moles={(12×2)+(1×6)+16} × 55.5556= 46.5×55.5556= 2555.55

4 0
2 years ago
Read 2 more answers
The constant volume heat capacity of a gas can be measured by observing the decrease in temperature when it expands adiabaticall
miv72 [106K]

Answer:

The value is  C_p  = 42. 8 J/K\cdot mol

Explanation:

From the question we are told that  

     \gamma = \frac{C_p }{C_v}

The  initial volume of the  fluorocarbon gas is  V_1 = V

 The final  volume of the fluorocarbon gas isV_2 = 2V

  The initial  temperature of the fluorocarbon gas is  T_1  =  298.15 K

  The final  temperature of the fluorocarbon gas is T_2  =  248.44 K

   The initial  pressure is P_1  = 202.94\  kPa

    The final   pressure is  P_2  =  81.840\  kPa

Generally the equation for  adiabatically reversible expansion is mathematically represented as

       T_2 =  T_1  * [ \frac{V_1}{V_2} ]^{\frac{R}{C_v} }

Here R is the ideal gas constant with the value  

        R =  8.314\  J/K \cdot mol

So  

   248.44 =   298.15  * [ \frac{V}{2V} ]^{\frac{8.314}{C_v} }

=> C_v  =  31.54 J/K\cdot mol

Generally adiabatic reversible expansion can also be mathematically expressed as

    P_2 V_2^{\gamma} = P_1 V_1^{\gamma}

=>[ 81.840 *10^3] [2V]^{\gamma} = [202.94 *10^3] V^{\gamma}    

=>  2^{\gamma} =  2.56

=>    \gamma =  1.356

So

     \gamma  =  \frac{C_p}{C_v} \equiv  1.356 = \frac{C_p}{31.54}

=>    C_p  = 42. 8 J/K\cdot mol

3 0
3 years ago
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