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ryzh [129]
3 years ago
13

Which substance is a base? HCOOH RbOH H2CO3 NaNO3

Chemistry
2 answers:
harkovskaia [24]3 years ago
7 0

Answer:

RbOH

Explanation:

Alex Ar [27]3 years ago
6 0

Answer:

RbOH

Explanation:

For this question, we have to remember what is the definition of a base. A base is a compound that has the <u>ability to produce hydroxyl ions</u> OH^-, so:

AOH~->~A^+~+~OH^-

With this in mind we can write the <u>reaction for each substance:</u>

HCOOH~->~HCOO^-~+~H^+

RbOH~->~Rb^+~+~OH^-

H_2CO_3~->~CO_3^-^2~+~2H^+

NaNO_3~->~Na^+~+~NO_3^-

The only compound that fits with the definition is RbOH, so this is our <u>base</u>.

I hope it helps!

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Nitric acid decomposes in light to form nitrogen dioxide, water, and oxygen, according to the following chemical equation.
geniusboy [140]

Answer:

b. 0.22 L

Explanation:

Step 1: Write the balanced equation

4 HNO₃(l) ⇒ 4 NO₂(g) + 2 H₂O(l) + O₂(g)

Step 2: Calculate the moles corresponding to 2.5 g of HNO₃

The molar mass of HNO₃ is 63.01 g/mol.

2.5 g × 1 mol/63.01 g = 0.040 mol

Step 3: Calculate the moles of O₂ produced from 0.040 moles of HNO₃

The molar ratio of HNO₃ to O₂ is 4:1. The moles of O₂ produced are 1/4 × 0.040 mol = 0.010 mol.

Step 4: Calculate the volume corresponding to 0.010 moles of O₂

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7 0
3 years ago
Student 1You are supplied with a 100 mL beaker that you weigh on a pocket balance – its weight clean and dry is 29.3 g. You then
navik [9.2K]

Answer:

The answer to your question is: density = 0.993 g/ml

Explanation:

Data

mass beaker empty = 29.3 g

volume of liquid = 15 ml

mass beaker + liquid = 44.2 g

Formula

             density = mass / volume

Process

mass of liquid = 44.2 - 29.3

                       = 14.9 g

             density = 14.9 / 15

                          = 0.993 g/ml

6 0
3 years ago
The compound consists of 40% carbon and 6.71% hydrogen. Its mass of 1 liter of steam is 2.4 g. What is the formula of the compou
AveGali [126]

I think it is CH2O.

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7 0
3 years ago
Calculate the equilibrium constant for the reaction: 2 Cr + 3 Pb2+ ----&gt; 3 Pb + 2 Cr3+ at 25oC. Eocell = 0.61 V
sattari [20]

Answer:

The value is  K  =  8*10^{61}

Explanation:

From the question we are told that

    The equation is  2 Cr  +  3Pb^{2+} \to 3Pb + 2Cr^{3+}

     The  temperature is  T = 25^oC =  298 K   [room  \ temperature ]

     The  emf at standard condition is  E^o_{cell}  =  0.61 \  V

Generally at the cathode

      3Pb^{2+}(aq) + 6 e- --> 3Pb(s)

  At the anode

      2Cr^{3+} + 6e^- \to  2Cr

Generally for an  electrochemical reaction, at room temperature the Gibbs free energy is mathematically represented as  

       G =  n*  F *  E^o_{cell}

Here  n  is  the no of electron  with value n = 6

       F  is  the Faraday's constant with value 96487 J/V

  =>   G =  6  * 96487 *  0.61

  =>   G = 3.5 *10^{5} \  J

This Gibbs free energy can also be represented mathematically as

       G =  RTlogK

Here  R  is the cell constant with value 8.314J/K

           K is the equilibrium constant

From above

=>  K  =  antilog^{\frac{G}{ RT} }

Generally  antilog =  2.718

=>K  =  2.718^{\frac{3.5 *10^5}{ 8.314* 298} }

=>   K  =  8*10^{61}

       

         

       

         

6 0
3 years ago
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