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Tomtit [17]
2 years ago
13

Sodium combines with water to produce sodium hydroxide and hydrogen gas. Which word equation represents this violent reaction?

Chemistry
2 answers:
Natalka [10]2 years ago
5 0

Answer: sodium + water → sodium hydroxide + hydrogen

Explanation:

1) The first part of the statment "sodium combines with water" means that these two pure substances are the reactants.

Sodium is an alkali metal (group 1: Li, Na, K, Rb, Cs, Fr).

A typical chemical property of the alkali metal elments is that they react violently with water fo form the corresponding hydroxide and hydrogen gas.

2) The next two words, "to produce ". indicates, precisely, that what follow is the product of the reaction. In the chemical reaction it is indicated by the arrow →

3) Finally, you are told the products of the reaction: sodium hydroxide and hydrogen gas.

4) The word equation then follows the order reactants → products, which in this case is the sodium + water → sodium hydroxide + hydrogen

5) Using chemical symbols and conventions, that is:

Na(s) + H₂O (l) → NaOH(aq) + H₂(g).

nataly862011 [7]2 years ago
5 0

Answer: sodium + water → sodium hydroxide + hydrogen

Explanation: Taking the quiz rn

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A 1.87 L aqueous solution of KOH contains 155 g of KOH . The solution has a density of 1.29 g/mL . Calculate the molarity ( M ),
lbvjy [14]

Answer:

[KOH] = 1.47 M

[KOH] = 1.22 m

KOH = 6.86 % m/m

Explanation:

Let's analyse the data

1.87 L is the volume of solution

Density is 1.29 g/mL → Solution density

155 g of KOH → Mass of solute

Moles of solute is (mass / molar mass) = 2.76 moles.

Molarity is mol/L → 2.76 mol / 1.87 L = 1.47 M

Let's determine, the mass of solvent.

Molality is mol of solute / 1kg of solvent

We can use density to find out the mass of solution

Mass of solution - Mass of solute = Mass of solvent

Density = Mass / volume

1.29 g/mL = Mass / 1870 mL

Notice, we had to convert L to mL, cause the units of density.

1.29 g/mL . 1870 mL = Mass → 2412.3 g

2412.3 g - 155 g = 2257.3 g of solvent

Let's convert the mass of solvent to kg

2257.3 g / 1000 = 2.25kg

2.76 mol / 2.25kg = 1.22 m (molality)

% percent by mass = mass of solute in 100g of solution.

(155 g / 2257.3 g) . 100g = 6.86 % m/m

5 0
2 years ago
In the formula X2O5, the symbol X could represent an element in Group
topjm [15]

Answer: (3) 15

Explanation: We criss-cross down the oxidation numbers to get the subscripts for the correct formulas. That means the X has an oxidation number of 5. The element with the + oxidation number is always written first so it is +5. Of the groups names, only group 15 has +5 as an oxidation number.

6 0
2 years ago
In Philip’s French class, the students are learning how to pronounce closed vowels and open vowels. The students are most likely
Sladkaya [172]

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It sounds like they are studying French phonemes

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7 0
2 years ago
Read 2 more answers
If 34.7 g of AgNO₃ react with 28.6 g of H₂SO₄ according to this UNBALANCED equation below, how many grams of Ag₂SO₄ could be for
Luba_88 [7]

The  number  of grams   of Ag2SO4  that could be formed  is   31.8  grams



    <u><em> calculation</em></u>

Balanced   equation is  as below

2 AgNO3 (aq)  + H2SO4(aq)  →  Ag2SO4 (s)   +2 HNO3 (aq)


  • Find  the  moles  of  each reactant by use  of  mole= mass/molar mass  formula

that is  moles of  AgNO3= 34.7 g / 169.87  g/mol= 0.204 moles

             moles of  H2SO4 =  28.6  g/98  g/mol  =0.292  moles

  • use the  mole  ratio to determine the moles of  Ag2SO4

   that is;

  •    the mole ratio of  AgNo3 : Ag2SO4 is  2:1 therefore  the  moles of Ag2SO4=  0.204  x1/2=0.102 moles

  • The moles  ratio of H2SO4  : Ag2SO4  is  1:1  therefore  the moles of Ag2SO4 = 0.292  moles

 

  •      AgNO3  is the limiting reagent therefore  the moles of   Ag2SO4 = 0.102  moles

<h3>     finally  find  the mass  of Ag2SO4  by use of    mass=mole  x molar mass  formula</h3>

that  is  0.102   moles  x  311.8  g/mol= 31.8 grams

3 0
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list three sources of error that could account for the differences between your values for the enthalpy of fusion of water and t
NARA [144]

<span>Three sources of error that might account for the differences in the enthalpy of fusion include the room temperature how much’ long you stirred and another thing that might make it have different results is how long the ice was out for   </span>


3 0
2 years ago
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