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nika2105 [10]
3 years ago
6

State the bond angle and the molecular shape of NH3 and H2O.​

Chemistry
1 answer:
FinnZ [79.3K]3 years ago
7 0

Answer:

H2O forms a bent structure due to the presence of 2 lone pairs over O Atom. This makes the bond angle 104.5° NH3 forms a pyramidal structure due to presence of 1 lone pair. This makes the bond angle 107°

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The maximum allowable concentration of pb2+ ions in drinking water is 0.05 ppm (i.e., 0.05 g of pb2+ in 1 million grams of water
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PbSO₄ partially dissociates in water. the balanced equation is;
                    
                       PbSO₄(s) ⇄  Pb²⁺(aq) + SO₄²⁻(aq)
Initial                                     -                -
Change             -X               +X           +X
Equilibrium                           X              X

Ksp           =    [Pb²⁺(aq)] [SO₄²⁻(aq)]
1.6 x 10⁻⁸  =    X * X
1.6 x 10⁻⁸  =    X²
          X    =   1.3 x 10⁻⁴ M
      
Hence the Pb²⁺ concentration in underground water is 1.3 x 10⁻⁴ M. 
[Pb²⁺]  = 1.3 x 10⁻⁴ M.
           = 1.3 x 10⁻⁴ mol / L x 207 g / mol 
           = 26.91 ppm

8 0
3 years ago
True/False. A program is an algorithm that has been coded into something that can be run by a machine. *
Zepler [3.9K]

Answer:

true

Explanation:

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3 years ago
Help i attached question
Oksanka [162]

Answer:

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6 0
3 years ago
Read 2 more answers
Water is denser than oil. When frozen, the oil sinks to the bottom. Why?
sammy [17]

Answer: When water freezes it gets larger and becomes a solid but it still weighs the same as when it was a liquid form. When it becomes larger it takes up space but makes it less dense. Ok so getting to the point- oil unlike water becomes more dense when frozen so this is why oil sinks in water. Sorry if this was confusing but I hope it helped have a great day and god bless you :3

6 0
3 years ago
7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c.
beks73 [17]

Answer:

\Delta H_{f,C_3H_4}=276.8kJ/mol

Explanation:

Hello!

In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

\Delta H_{rxn} =- m_wC_w\Delta T

We plug in the mass of water, temperature change and specific heat to obtain:

\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ

Now, this enthalpy of reaction corresponds to the combustion of propyne:

C_3H_4+4O_2\rightarrow 3CO_2+2H_2O

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol

Now, we solve for the enthalpy of formation of C3H4 as shown below:

\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol

Best regards!

7 0
3 years ago
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