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Aleonysh [2.5K]
3 years ago
9

How many centimeters are in 1 mile? Use the equalities 1 mile = 5280 feet; 1 foot = 12 inches; and 1 centimeter = 0.394 inches.

A. (5280 x 12 ∕ 0.394) cm B. 5280 ∕ (12 x 0.394) cm C. 5280 x 12 x 0.394 cm D. (5280 + 12 + 0.394) cm
Chemistry
2 answers:
Galina-37 [17]3 years ago
7 0
The answer is A which comes out to about 160812 cm
Alla [95]3 years ago
4 0

Answer:

A. \frac{5280*12}{0.394}cm

Explanation:

1. Take the quantity you want to convert:

1mile

2. Multiply by the first conversion factor:

1mile*\frac{5280feet}{1mile}

3. Apply the second conversion factor:

1mile*\frac{5280feet}{1mile}*\frac{12inches}{1foot}

4. Multiply by the third conversion factor:

1mile*\frac{5280feet}{1mile}*\frac{12inches}{1foot}*\frac{1centimeter}{0.394inches}

5. Multiply all the numerators and divide by denominators:

1mile*\frac{5280feet}{1mile}*\frac{12inches}{1foot}*\frac{1centimeter}{0.394inches}=\frac{5280*12}{0.394}cm

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shutvik [7]

Answer:

0.0917 mol Co(CrO₄)₃

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

37.3 g Co(CrO₄)₃

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

Molar Mass of Co - 58.93 g/mol

Molar Mass of Cr - 52.00 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Co(CrO₄)₃ - 58.93 + 3(52.00) + 12(16.00) = 406.93 g/mol

<u>Step 3: Convert</u>

<u />37.3 \ g \ Co(CrO_4)_3(\frac{1 \ mol \ Co(CrO_4)_3}{406.93 \ g \ Co(CrO_4)_3} ) = 0.091662 mol Co(CrO₄)₃

<u>Step 4: Check</u>

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

0.091662 mol Co(CrO₄)₃ ≈ 0.0917 mol Co(CrO₄)₃

8 0
2 years ago
A solution has a [OH-] of 1 × 10-9. What is the pOH of this solution?
zhannawk [14.2K]
POH = - log [ OH⁻ ]

pOH = - log [ 1 x 10⁻⁹ ] 

pOH = 9

Answer C

hope this helps!
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3 years ago
Who is telling what happens in Home Début?
nekit [7.7K]

Answer:

D

Explanation:

Its in first person

8 0
3 years ago
A bubble of helium gas has a volume of 0.65 L near the bottom of a large
Flura [38]

Answer:

0.73L

Explanation:

The following data were obtained from the question :

V1 = 0.65 L

P1 = 3.4 atm

T1 = 19°C = 19 + 273 = 292K

V2 =?

P2 = 3.2 atm

T2 = 36°C = 36 + 273 = 309K

The bubble's volume near the top can be obtain as follows:

P1V1 /T1 = P2V2 /T2

3.4 x 0.65/292 = 3.2 x V2 /309

Cross multiply to express in linear form as shown below:

292 x 3.2 x V2 = 3.4 x 0.65 x 309

Divide both side by 292 x 3.2

V2 = (3.4 x 0.65 x 309) /(292 x 3.2)

V2 = 0.73L

Therefore, the bubble's volume near the top is 0.73L

8 0
3 years ago
Balance this equation. If a coefficient of "1" is required, choose "blank" for that box.
Kitty [74]

Answer: 2 C2H4 + 6 O2 => 4 CO2 + 4 H2O

Explanation:The coefficient are as follows: 2: 6: 4: 4

Each atom on the reactant and product side are equal.

Reactant Product

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H 2x4 = 8 4x2 = 8

O 6x2 = 12 (4x2) + 4 = 12

7 0
3 years ago
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