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madreJ [45]
3 years ago
10

if the temperature increases and the barometric pressure Falls then which symbols will denote this on a weather map

Chemistry
2 answers:
ohaa [14]3 years ago
5 0

Warm, moist air mass is approaching as a warm front. Is the answer


Likurg_2 [28]3 years ago
5 0

The answer is, warm, moist air is approaching as a warm front.

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Bcz you’re able to wear something fresh, get a tan if you’d want, play volleyball or go out to swim in the cold ocean that feels so good when it’s hot !
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What is the molecular formula of a substance that decomposes into 1.33 g of h and 21.3 g of o, and was found to have a molar mas
Tema [17]
<span>H2O2 First, let's determine how many moles of hydrogen and oxygen atoms we have. Start by looking up the atomic weights of those elements: Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Moles hydrogen = 1.33 g / 1.00794 g/mol = 1.319522987 mol Moles oxygen = 21.3 g / 15.999 g/mol = 1.331333208 mol We now have a ratio of 1.319522987 : 1.331333208 and we want a ratio of small integers that is close. Start by dividing all the numbers in the ratio by the smallest value, giving: 1 : 1.008950371 This ratio is acceptably close to 1:1 so I assume the formula is of the form HnOn where n is a small integer. Let's initially assume that n is 1, so the mass would be 1.00794 + 15.999 = 17.00694 Obviously 17 is far smaller than 34.1. So let's divide 34.1 by 17.00694 and see what n should be: 34.1 / 17.00694 = 2.005063815 So the formula we want is H2O2, which is hydrogen peroxide.</span>
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When 1.025 g of naphthalene (c10h8) burns in a bomb calorimeter, the temperature rises from 24.25°c to 32.33°c. find δerxn for t
Goryan [66]

Answer : The \Delta E for the combustion of naphthalene is 5161.25KJ/mole

Solution : Given,

Mass of naphthalene = 1.025 g

Initial temperature = 24.25^oC

Final temperature = 32.33^oC

Specific heat capacity of calorimeter = 5.11KJ/^oC

Molar mass of naphthalene = 128 g/mole

First, we have to calculate the heat absorbed, q

Formula used :

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

Now put all the given values in this formula, we get

q=(5.11KJ/^oC)\times (32.33^oCT-24.25^oC)=41.29KJ

Now we have to calculate the moles of naphthalene.

Moles of C_{10}H_{8} = \frac{\text{ Mass of }C_{10}H_{8}}{\text{ Molar mass of }C_{10}H_{8}}=\frac{1.025g}{128g/mole}=0.0080moles

Now we have to calculate the \Delta E for combustion of naphthalene.

\Delta E=\frac{q}{n}

where,

q = heat absorbed

n = number of moles

Now put all the values in this formula, we get

\Delta E=\frac{41.29KJ}{0.0080moles}=5161.25KJ/mole

Therefore, the \Delta E for the combustion of naphthalene is 5161.25KJ/mole

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Convert 175 lbs to kg (1 kg = 2.2 lbs)
ankoles [38]

Answer:

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Explanation:

Divide 175 by 2.2

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