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When the chemical equation is balanced what is the ratio of the coefficient of Ag(s) to the cofficient of H2O(l)?___Ag(s)+____NO3-(aq)+____H+(aq)→ ___Ag+(aq)+____NO(g)+______H2O(l)a. 1:1b. 2:1c. 3:1d. 3:2

Answers

Answer 1
Answer:

Answer:

Ag(s):H2O(l)  = 3:2

For 3 moles Ag(s) we'll have 2 moles H2O(l)

Option D is correct

Explanation:

Step 1: Balancing the equation

3 Ag (s) + 4 HNO3 (aq) → 3 AgNO3 (aq) + NO (g) + 2 H2O (l)

3Ag(s)  +  4NO  ^3-  +  4H+  →3Ag+ +3NO3- +  +NO  +  2H2O

3Ag(s)  +  NO  ^3-(aq)  +  4H+(aq)  →3Ag+(aq)  +NO(g)  +  2H2O(l)

Step 2: The ratio between Ag(s) and H2O(l)

Ag(s):H2O(l)  = 3:2

For 3 moles Ag(s) we'll have 2 moles H2O(l)

Option D is correct

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Barium chloride + sodium phosphate 
Answer with double replacement

Answers

Answer:

Barium chloride + Sodium phosphate    →   barium phosphate + sodium chloride

Explanation:

Double replacement:

It is the reaction in which two compound exchange their ions and form new compounds.

AB + CD → AC +BD

Chemical equation:

BaCl₂ + Na₃PO₄         →     Ba₃(PO₄)₂ + NaCl

Balanced chemical equation:

3BaCl₂ + 2Na₃PO₄         →     Ba₃(PO₄)₂ + 6NaCl

The cation and anion of both reactants are exchanged with each other.

Ba²⁺ react with PO₄³⁻ and form Ba₃(PO₄)₂ while Cl⁻ react with Na⁺ and form sodium chloride.  

Molecular equation:

Barium chloride + Sodium phosphate    →   barium phosphate + sodium chloride

Final answer:

A double replacement reaction between barium chloride and sodium phosphate results in the formation of barium phosphate and sodium chloride.

Explanation:

The given chemical equation represents a double replacement reaction. The chemical reactants are barium chloride (BaCl2) and sodium phosphate (Na3PO4). In a double replacement reaction, the cations and anions of the two reactants switch places to form two new compounds.

So here is how the reaction would proceed: BaCl2 + Na3PO4 -> Ba3(PO4)2 + NaCl.

This translates to: Barium chloride reacts with sodium phosphate to form barium phosphate and sodium chloride.

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What is the empirical formula of a compound composed of 36.7 g 36.7 g of potassium ( K K ) and 7.51 g 7.51 g of oxygen ( O O )? Insert subscripts as needed.

Answers

Answer:

K₂O

Explanation:

Given parameters:

Mass of K = 36.7g

Mass of O = 7.51g

Unknown:

Empirical formula of the compound

Solution:

The empirical formula of a compound is it's simplest ratio by which the elements in the compound combines. It differs from the molecular formula that shows the actual atomic ratios.

To find the empirical formula, follow this process;

   

Elements                     K                               O

Mass                          36.7                            7.51

Molar

mass                            39                            16

Number of

moles                       36.7/39                    7.51/16

                                   0.94                        0.47

Divide by

the smallest              0.94/0.47               0.47/0.47

                                         2                              1

  Empirical formula is        K₂O

                                     

Final answer:

The empirical formula of the compound composed of 36.7 g of potassium and 7.51 g of oxygen is K2O.

Explanation:

To determine the empirical formula of a compound, we need to find the ratio of the elements present. In this case, we have 36.7 g of potassium and 7.51 g of oxygen. To find the ratio, we need to convert these masses to moles by dividing them by the molar masses of potassium and oxygen. The molar mass of potassium is 39.10 g/mol and the molar mass of oxygen is 16.00 g/mol. Dividing the masses by the molar masses gives us 0.939 mol potassium and 0.469 mol oxygen. The ratio between these two elements is approximately 2:1, so the empirical formula of the compound is K2O.

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An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution? Kb = 0.51°C/m for water. Enter your answer with 2 decimal places and no units.

Answers

Answer: The boiling point of solution is 100.53

Explanation:

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

\text{Boiling point of solution}-\text{Boiling point of pure solution}=i* K_b* \frac{m_(solute)* 1000}{M_(solute)* W_(solvent)\text{ (in grams)}}

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

K_b = molal boiling point elevation constant = 0.51°C/m

m_(solute) = Given mass of solute (CsCl) = 8.00 g

M_(solute) = Molar mass of solute (CsCl) = 168.4  g/mol

W_(solvent) = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

\text{Boiling point of solution}-100=2* 0.51^oC/m* (8.00* 1000)/(168.4g/mol* 92)\n\n\text{Boiling point of solution}=100.53^oC

Hence, the boiling point of solution is 100.53

What is the most likely position for the hurricane indicated by the wind readings from the three weather stations shown?

Answers

Here's the answer, I remember doing this problem last year.

23.5 degrees north, 77 degrees west

What is a characteristic of property acids pls help

Answers

Answer:

General Characteristics of Acids:

Sour taste (though you should never use this characteristic to identify an acid in the lab) Reacts with a metal to form hydrogen gas. Increases the H+ concentration in water. Donates H+ ions. Turns blue litmus indicator red.
ph>7
also turns blue litmus red

Which step would help a student find the molecular formula of a compound from the empirical formula? Multiply the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound. Subtract the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound from the subscripts of the empirical formula. Divide the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound. Add the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound to the subscripts of the empirical formula..

Answers

Answer:

Multiply the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound.

Explanation:

got it right on edge 2020 :)

Answer:

Multiply the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound.

Explanation:
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