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olga2289 [7]
3 years ago
6

If the mass of an object is 6 lbm, what is its weight, in lbf, at a location where g = 32.0 ft/s2?

Chemistry
1 answer:
Solnce55 [7]3 years ago
8 0

Using formula:

w = m\times g   -(1)

where w is weight of object,

m is mass of the object, and

g is the acceleration of gravity.

mass, m = 6 lbm (given)

acceleration of gravity, g = 32.0 ft/s^{2} (given)

Now, substituting the values in equation (1):

w = 6 lbm \times 32 ft/s^{2} = 192 lbmft/s^{2}

Since, 1 lbf = 32.174 lbm ft/s^{2}

So, w = 192\times \frac{1}{32.174 lbf}

w = 5.976 lbf

Hence, the mass of an object is 5.976 lbf.

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Calculate the number of O atoms in 0.364 g of CaSO4 · 2H2O
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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:

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Converting grams to moles:

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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>

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3 years ago
"46.7 g of water at 80.6 oC is added to a calorimeter that contains 45.33 g of water at 40.6 oC. If the final temperature of the
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<u>Answer:</u> The specific heat of calorimeter is 30.68 J/g°C

<u>Explanation:</u>

When hot water is added to the calorimeter, the amount of heat released by the hot water will be equal to the amount of heat absorbed by cold water and calorimeter.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[(m_2\times c_2)+c_3](T_{final}-T_2)       ......(1)

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m_2 = mass of cold water = 45.33 g

T_{final} = final temperature = 59.4°C

T_1 = initial temperature of hot water = 80.6°C

T_2 = initial temperature of cold water = 40.6°C

c_1 = specific heat of hot water = 4.184 J/g°C

c_2 = specific heat of cold water = 4.184 J/g°C

c_3 = specific heat of calorimeter = ? J/g°C

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46.7\times 4.184\times (59.4-80.6)=-[(45.33\times 4.184)+c_3](59.4-40.6)

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