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egoroff_w [7]
3 years ago
11

10 points!! Please give me work with it and I will mark as brainlist.

Chemistry
1 answer:
Georgia [21]3 years ago
8 0

Answer: The molecular formula will be C_2H_4O_2

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 40.0 g

Mass of O = 53.3 g

Mass of H = 6.66 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40.0g}{12g/mole}=3.33moles

Moles of O =\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.3g}{16g/mole}=3.33moles

Moles of H =\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.66g}{1g/mole}=6.66moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{3.33}{3.33}=1

For O =\frac{3.33}{3.33}=1

For H = \frac{6.66}{3.33}=2

The ratio of C : O : H = 1: 1: 2

Hence the empirical formula is COH_2

The empirical weight of COH_2 = 1(12)+1(16)+2(1)= 30g.

The molecular weight = 60 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{60}{30}=2

The molecular formula will be=2\times CH_2O=C_2H_4O_2

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Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

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where,

c_1 = specific heat of iron =  560 J/(kg.K)

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T_f = final temperature of water and iron = ?

T_1 = initial temperature of iron = 352^oC=273+352=625K

T_2 = initial temperature of water = 20^oC=273+20=293K

Now put all the given values in the above formula, we get:

(825\times 10^{-3}kg)\times 560J/(kg.K)\times (T_f-625K)=-(40\times 10^{-3}kg)\times 4186J/(kg.K)\times (T_f-293K)

T_f=537.12K

Therefore, the final equilibrium temperature of the water and iron is, 537.12 K

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3 years ago
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Using equation:

V_2 = \frac{P_1 X V_1 XT_2}{P_2 X T_1}

V_2 =\frac{2 atm  X 20 ml X200 K}{4 atm  X 300 K}

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How many atoms are in a sample of chromium with a mass of 31 grams? O a 2.4 x 1024 atoms of chromium b 3.6 x 1023 atoms of chrom
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Answer:

b. 3.66x10²³ atoms of chromium.

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Then we <u>calculate how many atoms are there in 0.608 moles</u>, using <em>Avogadro's number</em>:

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