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irinina [24]
3 years ago
14

When was George Washington born

Chemistry
2 answers:
ASHA 777 [7]3 years ago
8 0

he was born on feb. 22, 1723

svetoff [14.1K]3 years ago
8 0

George Washington was born on February 22, 1732 in Westmoreland country, Virginia.

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2 NaOH (s) + CO2(g) → Na2CO3 (s) + H20 (I)
Paha777 [63]
<h3>Answer:</h3>

16.7 g H₂O

<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>Chemistry</u>

<u>Stoichiometry</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)

[Given] 1.85 mol NaOH

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

[RxN] 2 mol NaOH → 1 mol H₂O

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                               \displaystyle 1.85 \ mol \ NaOH(\frac{1 \ mol \ H_2O}{2 \ mol \ NaOH})(\frac{18.02 \ g \ H_2O}{1 \ mol \ H_2O})
  2. Multiply/Divide:                 \displaystyle 16.6685 \ g \ H_2O

<u>Step 4: Check</u>

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

16.6685 g H₂O ≈ 16.7 g H₂O

6 0
3 years ago
Which of the following formulas has the smallest percent
KiRa [710]

Answer: CF4

Explanation:

Calculate the molar mass of each compound. Divide the molar mass of Carbon by the molar mass of each compound, then multiply the answer by 100 to get the percentage.

CF4= 12+(19X 4)

=12+76= 88 g/mol

%C= 12/88 x 100= 13.64%

CO2= 12+(16 X 2)

12+32= 44 G/MOL

%C= 12/44 x 100= 27.3%

CH4= 12+ (1 X4)

=12+4

=16 G/MOL

%C= 12/16  X 100= 75%

C204

(12X2) + (16X4)

24+64

= 88 g/mol

%C= 24/88 x 100

= 27.3%

5 0
3 years ago
Ecstasy is one example of which of the following?
yarga [219]

Answer:

Narcotics

I think ans is

7 0
3 years ago
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juin [17]
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7 0
3 years ago
Consider a galvanic cell based on the reaction Al^3+_(aq) + Mg_(s) rightarrow Al_(s) + Mg^2+ _(aq) The half-reactions are Al^3+
grin007 [14]

<u>Answer:</u> The standard cell potential of the cell is -0.71 V

<u>Explanation:</u>

The half reactions follows:

<u>Oxidation half reaction:</u>  Mg\rightarrow Mg^{2+}+2e^-;E^o_{Mg^{2+}/Mg}=-2.37V  ( × 3)

<u>Reduction half reaction:</u>  Al^{3+}(aq.)+3e^-\rightarrow Al(s);E^o_{Al^{3+}/Al}=-1.66V  ( × 2)

The balanced cell reaction follows:

2Al^{3+}(aq.)+3Mg(s)\rightarrow 2Al(s)+3Mg^{2+}(aq.)

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Putting values in above equation, we get:

E^o_{cell}=-2.37-(-1.66)=-0.71V

Hence, the standard cell potential of the cell is -0.71 V

8 0
3 years ago
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