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8090 [49]
3 years ago
6

Three different samples were weighed using a different type of balance for each sample. The three were found to have masses of 0

.6160959 kg, 3.225 mg, and 5480.7 g. The total mass of the samples should be reported as?
Chemistry
1 answer:
Lena [83]3 years ago
3 0

Answer:

6.096799125kg

Explanation:

According to the question, three different samples weighed using different types of balance had masses: 0.6160959 kg, 3.225 mg, and 5480.7 g.

Based on observation, the mass units in the three measurements are different but must be uniform in order to find the total mass. Hence, we need to convert to the standard unit (S.I unit of mass), which is kilograms (kg)

Since 1kg equals 1,000,000mg

Hence, 3.225mg will be 3.225/1000000

= 0.000003225kg

Also, 1kg equals 1000g

Hence, 5480.7g will be 5480.7/1000

= 5.4087kg

Hence, the total mass of the three samples (now in the same unit) are:

5.4807kg + 0.000003225kg + 0.6160959 kg

= 6.096799125kg

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<u>Answer:</u>

(A)

Density = Mass / Volume

So  

Mass = Density × Volume

= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene

1C_6 H_5 CH_3  + 9 O_2  > 7 CO_2  + 4 H_2 O

Mole ratio of toluene : Oxygen is 1 : 9

$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$

(B)

1 mole of Toluene produces 7 moles of CO_2 gas and 4 moles of H_2 O Vapour

So the mole ratio is 1 : 11

37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} $$\\\\=4.49 \text { mol gaseous products (Answer) } $

(C)

1mole contains 6.022\times10^{23} molecules

37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene}} \times \frac{4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \text { toluene }} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} $\\\\$=9.82 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O} \text { (Answer) } $

6 0
3 years ago
Which statement indicates that a substance may be an ionic compound?
lapo4ka [179]
If it has a metal and a nonmetal. Not sure if this helps, but I hope it does :)
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3 years ago
a thermometer containing 8.3g of mercury has broken. if mercury ha a density of 13.6g/mL. what volume is spilled?
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V = 8.3 / 13.6

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hope this helps!
4 0
3 years ago
Read 2 more answers
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multiply each # of moles times the atomic molar mass of the compund to find the relation is weights

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You cannot obtain 72 g of CO2 from 18 g of C.

May be they just pretended that you use the law of consrvation of mass and say that you need 72 g - 18g = 54 g. But it violates the proportion of C and O2 in the CO2 and is not possible.
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