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andrew-mc [135]
3 years ago
5

Match the reaction on the right with the corresponding reaction on the left by writing the correct, matching number in the space

provided. solid sodium + chlorine gas → solid sodium chloride solid sodium + liquid water → aqueous sodium hydroxide + hydrogen gas 2H2(g) + O2(g) → 2H2O(g) 2H2O2(aq) → O2(g) + 2H2O(l) 1) aqueous hydrogen peroxide → oxygen gas + liquid water 2) 2Na(s) + Cl2(g) → 2NaCl(s) 3) hydrogen gas + oxygen gas → water gas 4) 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Chemistry
2 answers:
Vesna [10]3 years ago
8 0

Explanation:

The first given equation in words is as follows.

solid sodium + chlorine gas \rightarrow solid sodium chloride

The chemical equation for this reaction will be written as follows.

        Na(s) + Cl(g) \rightarrow NaCl (s)

The second given equation in words is as follows.

solid sodium + liquid water \rightarrow aqueous sodium hydroxide + hydrogen gas

The chemical equation for this reaction will be written as follows.

    Na(s) + H_{2}O(l) \rightarrow NaOH(aq) + H_{2}

The third given equation is as follows.

    2H_{2}(g) + O_{2}(g) \rightarrow 2H_{2}O(g)

This equation can be written in words as follows.

hydrogen gas + oxygen gas \rightarrow water gas

The fourth given equation is as follows.

     2H_{2}O_{2}(aq) \rightarrow O_{2}(g) + 2H_{2}O(l)

This equation can be written in words as follows.

aqueous hydrogen peroxide \rightarrow oxygen gas + liquid water

Thus, match the given data as follows.

  • solid sodium + chlorine gas \rightarrow solid sodium chloride - (2).  2Na(s) + Cl_{2}(g) \rightarrow 2NaCl (s)
  • solid sodium + liquid water \rightarrow aqueous sodium hydroxide + hydrogen gas - (4).  2Na(s) + 2H_{2}O(l) \rightarrow 2NaOH(aq) + H_{2}
  • 2H_{2}(g) + O_{2}(g) \rightarrow 2H_{2}O(g) - (3). hydrogen gas + oxygen gas \rightarrow water gas.
  • 2H_{2}O_{2}(aq) \rightarrow O_{2}(g) + 2H_{2}O(l) - (1). aqueous hydrogen peroxide \rightarrow oxygen gas + liquid water


kiruha [24]3 years ago
8 0

Answer:

The answers are

2

4

3

1

Explanation:

they're all correct

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If 10.0 grams of NaHCO3 is added to 10.0 g of HCl, determine the efficiency of baking soda as an antacid if 6.73 g of NaCl was p
Lapatulllka [165]

Answer:

percentage yield of NaCl = 96.64%

Explanation:

The reaction was between NaHCO3 and HCl .The chemical equation can be represented below:

NaHCO3 + HCl → NaCl + H2O + CO2 . The balance equation is

NaHCO3 + HCl → NaCl + H2O + CO2

The question ask us to calculate the percentage yield of NaCl.

The efficiency of NaHCO3 as an antacid , the limiting reactant is NaHCO3

as

1 mole of NaHCO3 produces 1 mole of NaCl

Therefore,

molar mass of NaHCO3 = 23 +1 + 12 + 48 = 84 g

molar mass of NaCl = 23 + 35.5 = 58.5 g

1 mole of NaHCO3 = 84 g

1 mole of NaCl  = 58.5 g

since 84 g of NaHCO3 produces 58.5 g of NaCl

10 g of NaHCO3 will produce ? grams of NaCl

cross multiply

Theoretical yield of NaCl = (10 × 58.5)/84

Theoretical yield of NaCl = 585/84

Theoretical yield of NaCl  = 6.9642857143 g

percentage yield of NaCl = actual yield/theoretical yield × 100

percentage yield of NaCl = 6.73/6.9642857143 × 100

percentage yield of NaCl = 673/6.9642857143

percentage yield of NaCl = 96.635897436%

percentage yield of NaCl = 96.64%

3 0
3 years ago
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Can you force the atoms into new configurations by pushing atoms around
marta [7]
<span>No, this is not possible. The reason for this impossibility is that the electrons are not attracted to one another, and thus, would repel. If you were to push the atoms into each other, the electrons would then prevent this from happening. Essentially atoms will bounce off of one another due to the repel of charges.</span>
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3 years ago
How many moles are in 1.42 x 10^25<br> molecules of NaCl?
Sholpan [36]
<h3>Answer:</h3>

23.6 mol NaCl

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

1.42 × 10²⁵ molecules NaCl

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                             \displaystyle 1.42 \cdot 10^{25} \ molecules \ NaCl(\frac{1 \ mol \ NaCl}{6.022 \cdot 10^{23} \ molecules \ NaCl})
  2. Multiply/Divide:                                                                                                   \displaystyle 23.5802 \ mol \ NaCl

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

23.5802 mol NaCl ≈ 23.6 mol NaCl

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3 years ago
The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 60.21°C? (ΔHvap for ethanol is 39.3 k
Delicious77 [7]

Answer:

The vapor pressure at 60.21°C is 327 mmHg.

Explanation:

Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.

We need to find vapor pressure at 60.21°C.

The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.

ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})

We have given in the question

P_1=102\ mmHg

T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol

And R is the Universal Gas Constant.

R=0.008 314 kJ/Kmol

ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165

Taking inverse log both side we get,

\frac{P_2}{102}=e^{1.165}\\\\P_2=102\times 3.20\ mmHg\\P_2=327\ mmHg

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