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2 years ago

216 grams of carbon 36 grams of hydrogen and 288 grams of oxygen whats the empirical formula

1 answer:

4 0

Divide each wight by the relative atomic mass

C = 216 / 12 = 18
H = 36 / 1 = 36
O = 288/16 = 18

Ratio of C:h:O = 1:2:1

Empirical formula is CH2O Could be formaldehyde HCHO.

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What is the % yield when 12g of Mg react with HCl to produce 0.2g of H2?

goblinko [34]

Answer:

The yield is 20 %.

Explanation:

Mg + 2 HCl → MgCl2 + H2

(12 g Mg) / (24.30506 g Mg/mol) x (1 mol H2 / 1 mol Mg) x (2.015894 g H2/mol) = 0.9953 g H2.

(0.2 g) / (0.9953 g) = 0.20 = 20% yield

5 0

2 years ago

553.5 mL ?:cm^3<br> how do i get my answer? and what is the answer?

dmitriy555 [2]

Answer:

553.5 cm3

Reason:

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6 0

3 years ago

Suppose 550.mmol of electrons must be transported from one side of an electrochemical cell to another in 49.0 minutes. Calculate

nekit [7.7K]

Answer:

18.0 Ampere is the size of electric current that must flow.

Explanation:

Moles of electron , n = 550 mmol = 0.550 mol

1 mmol = 0.001 mol

Number of electrons = N

$N=N_A\times n$

Charge on N electrons : Q

$Q = N\times 1.602\times 10^{-19} C$

Duration of time charge allowed to pass = T = 49.0 min = 49.0 × 60 seconds

1 min = 60 seconds

Size of current : I

$I=\frac{Q}{T}=\frac{N\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}$

$=\frac{n\times N_A\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}$

$I=\frac{0.550 mol\times 6.022\times 10^{23} mol^{-1}\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}=18.047 A\approx 18.0 A$

18.0 Ampere is the size of electric current that must flow.

3 0

3 years ago

Why does a balloon pop when it is stepped on?

KiRa [710]

Answer:

B. The pressure has increased

Explanation:

6 0

2 years ago

a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding empirical formula is

fiasKO [112]

Answer: The empirical formula is $CH_2$ .

Explanation:

Mass of C = 1.71 g

Mass of H = 0.287 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{1.71g}{12g/mole}=0.142moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.287g}{1g/mole}=0.287moles$

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

For C = $\frac{0.142}{0.142}=1$

For H = $\frac{0.287}{0.142}=2$

The ratio of C: H = 1: 2

Hence the empirical formula is $CH_2$ .

4 0

3 years ago