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vampirchik [111]
3 years ago
7

The pressure inside a tire is measured as 28.0 . What is its pressure in ? 1 pound = 4.45 newtons 1 inch2 = 6.45 centimeters2 Ex

press the answer to the correct number of significant figures. The pressure is .
Chemistry
2 answers:
Zigmanuir [339]3 years ago
7 0

Given: The pressure inside a tire (P) = 28.0 psi = 28.0 pound / inch²

And 1 pound = 4.45 newtons 1 inch² = 6.45 cm²

Calculation: We shall convert the given pressure in terms of Nm⁻²

Since,  

1 pound / inch² =  [1 pound× ( 4.45 N / 1 pound)] / [1 inch²× ( 6.45 cm² /1 inch²)]

or,                   =  [1 pound× ( 4.45 N / 1 pound)] / [1 inch²× ( 6.45×10⁻⁴ m² /1 inch²)]

or,                   =  6.899 ×10³ Nm⁻²

The pressure inside a tire (P) = 28.0 psi

or,                                            = 28.0 pound / inch²

or,                                            = 28.0 × 6.899 ×10³ Nm⁻²

or,                                            = 193.172 ×10³ Nm⁻²

or,                                            = 1.93172 ×10⁵ Nm⁻²

Since the given measurements have three significant figures.

Hence, the required pressure in the correct significant figures ( three digits) will be 1.93 ×10⁵ N m⁻².

Alisiya [41]3 years ago
5 0

Answer:

19.3

Explanation:

Answer on Plato

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4 0
3 years ago
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4 0
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A liquid occupies a volume of 5.0 L has a mass of 6.0 kg. what is the density of a the lights in kg/L
grandymaker [24]

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Hey there! :

Volume = 5.0 L

mass = 6.0 Kg

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6 0
3 years ago
If 3.31 moles of argon gas occupies a volume of 100 L what volume does 13.15 moles of argon occupy under the same temperature an
kumpel [21]

Answer:

397 L

Explanation:

Recall the ideal gas law:

\displaystyle PV = nRT

If temperature and pressure stays constant, we can rearrange all constant variables onto one side of the equation:

\displaystyle \frac{P}{RT} = \frac{n}{V}

The left-hand side is simply some constant. Hence, we can write that:

\displaystyle \frac{n_1}{V_1} = \frac{n_2}{V_2}

Substitute in known values:

\displaystyle \frac{(3.31 \text{ mol})}{(100 \text{ L})}  = \frac{(13.15\text{ mol })}{V_2}

Solving for <em>V</em>₂ yields:

\displaystyle V_2 = \frac{(100 \text{ L})(13.15)}{3.31} = 397 \text{ L}

In conclusion, 13.15 moles of argon will occupy 397* L under the same temperature and pressure.

(Assuming 100 L has three significant figures.)

3 0
2 years ago
Calculate the number of O atoms in 0.364 g of CaSO4 · 2H2O
Nikolay [14]

Answer:

<em>= 7.66 x 10²¹  oxygen atoms in 0.364 grams of  CaSO₄·2H₂O</em>

Explanation:

For problems like this posting, one needs an understanding of the following topics:

The definition of the mole

<u>1 mole of substance</u> = mass in grams of substance containing 1 Avogadro's Number ( = 6.023 x 10²³ ) of particles of the specified substance. This is generally one formula weight of the substance of interest. From this, the following equivalent relationships should be memorized:

<em>   1 mole = 1 formula weight = 1 mole weight (g)= 6.023 x 10²³ particles</em>

Converting grams to moles:

<em>Given grams => moles = grams/gram formula wt </em>

Converting moles to grams:

<em>Given moles => grams = moles x gram formula wt</em>

_________________________________________________________

<em>Calculate the number of O atoms in 0.364 g of CaSO₄ · 2H₂O.</em>

Given mass CaSO₄ · 2H₂O = 0.364 grams

Formula Wt CaSO₄ · 2H₂O = 172 g/mole

moles CaSO₄ · 2H₂O = mass <em>CaSO4 · 2H2O / formula Wt. CaSO₄ · 2H₂O</em>

<em>= 0.364 g CaSO₄·2H₂O </em><em>/ </em><em>172 g CaSO4·2H2O </em>

<em>= (0.364/172) mole CaSO₄·2H₂O </em>

<em>= 2.12 x 10⁻³ mole CaSO₄·2H₂O    </em>

<em>∴ number of Oxy (O) atoms in 0.364 grams CaSO₄·2H₂O </em>

<em>=  (2.12 x 10⁻³ mole CaSO₄ · 2H₂O)(6.023 x 10²³ molecules CaSO₄· 2H₂O/ mole)</em>

<em>= 1.276876 x 10²¹molecules CaSO₄· 2H₂O  CaSO₄2H₂O </em>

<em>= 1.276876 x 10²¹ molecules CaSO₄· 2H₂O   x   6 oxygen atoms / molecule</em>

<em>= 7.661256 x 10²¹  oxygen atoms in 0.364 grams of  CaSO₄·2H₂O</em>

<em>= 7.66 x 10²¹  oxygen atoms in 0.364 grams of  CaSO₄·2H₂O</em>

<em />

8 0
2 years ago
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