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3 years ago

How many joules of heat are required to heat 110 g of aluminum from 52.0 oC to 91.5 oC?

Chemistry

1 answer:

docker41 [41]3 years ago

6 0

Answer:

We need 3910.5 joules of energy

Explanation:

Step 1: Data given

Mass of aluminium = 110 grams

Initial temperature = 52.0 °C

Final temperature = 91.5 °C

Specific heat of aluminium = 0.900 J/g°C

Step 2: Calculate energy required

Q = m*c*ΔT

⇒with Q = the energy required = TO BE DETERMINED

⇒with m = the mass of aluminium = 110 grams

⇒with c = the specific heat of aluminium = 0.900 J/g°C

⇒with ΔT = the change in temperature = T2 - T1 = 91.5 °C - 52.0 °C = 39.5 °C

Q = 110 grams * 0.900 J/g°C * 39.5

Q = 3910.5 J

We need 3910.5 joules of energy

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Calculate the mass of magnesium carbonate ( MgCO3), in grams, required to produce 110.0 g of carbon dioxide using the following

bearhunter [10]

Answer:

$210.7~g~MgCO_3$

Explanation:

We have to start with the reaction:

$MgCO_3~->~MgO~+~CO_2$

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the number of moles that we have in the 110.0 g of carbon dioxide, to this, we have to know the atomic mass of each atom:

C: 12 g/mol

O: 16 g/mol

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If we take into account the number of atoms in the formula, we can calculate the molar mass of carbon dioxide:

$(12*1)+(16*2)=44~g/mol$

In other words: $1~mol~CO_2=~44~g~CO_2$ . With this in mind, we can calculate the moles:

$110~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=25~mol~CO_2$

Now, the molar ratio between carbon dioxide and magnesium carbonate is 1:1, so:

$2.5~mol~CO_2=2.5~mol~MgCO_3$

With the molar mass of $MgCO_3$ ( $(23.3*1)+(12*1)+(16*3)=84.3~g/mol$ . With this in mind, we can calculate the grams of magnesium carbonate:

$2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3$

I hope it helps!

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MAVERICK [17]

Answer:

shorter wavelength = alpha wave

Explanation:

Given that,

The alpha wave has a frequency of 5 Hz and the beta wave has a frequency of 2 Hz.

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For alpha wave,

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For beta wave,

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From the above calculations, we find that the wavelength of the alpha wave is shorter than the wavelength of the beta wave.

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~Hello there! ^_^

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Happy Studying! =)

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