All categories

2 years ago

What is the activity of a 52.3 μCi sample of carbon‑14 in becquerels?

Chemistry

1 answer:

Llana [10]2 years ago

3 0

Answer:

1935100 Bq

Explanation:

Let us recall that:

If 1 μCi can be equivalent to 37000 Bq

Then; the activity of 52.3 μCi will be:

$\dfrac{37000 \ Bq}{1 \ \mu Ci}\times 52.3 \ \mu Ci \\ \\ \\ \\ \mathbf{= 1935100 \ Bq}$

You might be interested in

Calculate the answer. Express it in scientific notation and include the correct number of significant figures. (12 x 104 ) x (5

ipn [44]

Answer:

Explanation:

(12 x 104 ) x (5 x 10-²) = 6 x 10 ³ 6 x 105 6.0x10²

1. (12 x 104 ) = 1248 or 1.248 x 10^3

2. (1.248 x 10^3)(5 x 10^-2) = 6.240 x 10^1 or 60 rounded to 1 sig fig

I don't understand "= 6 x 10 ³ 6 x 105 6.0x10² "

6 0

1 year ago

A gas contains 75.0 wt% methane, 10.0% ethane, 5.0% ethylene, and the balance water. (a) Calculate the molar composition of this

NeTakaya

Answer:

a) molar composition of this gas on both a wet and a dry basis are

5.76 moles and 5.20 moles respectively.

Ratio of moles of water to the moles of dry gas =0.108 moles

b) Total air required = 68.51 kmoles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

Explanation:

Let assume we have 100 g of mixture of gas:

Given that :

Mass of methane =75 g

Mass of ethane = 10 g

Mass of ethylene = 5 g

∴ Mass of the balanced water: 100 g - (75 g + 10 g + 5 g)

Their molar composition can be calculated as follows:

Molar mass of methane $CH_4}= 16 g/mol$

Molar mass of ethane $C_2H_6= 30 g/mol$

Molar mass of ethylene $C_2H_4 = 28 g/mol$

Molar mass of water $H_2O=18g/mol$

number of moles = $\frac{mass}{molar mass}$

Their molar composition can be calculated as follows:

$n_{CH_4}= \frac{75}{16}$

$n_{CH_4}=$ 4.69 moles

$n_{C_2H_6} = \frac{10}{30}$

$n_{C_2H_6} =$ 0.33 moles

$n_{C_2H_4} = \frac{5}{28}$

$n_{C_2H_4} =$ 0.18 moles

$n_{H_2O}= \frac{10}{18}$

$n_{H_2O}=$ 0.56 moles

Total moles of gases for wet basis = (4.69 + 0.33 + 0.18 + 0.56) moles

= 5.76 moles

Total moles of gas for dry basis = (5.76 - 0.56)moles

= 5.20 moles

Ratio of moles of water to the moles of dry gas = $\frac{n_{H_2O}}{n_{drygas}}$

= $\frac{0.56}{5.2}$

= 0.108 moles

b) If 100 kg/h of this fuel is burned with 30% excess air(combustion); then we have the following equations:

$CH_4 + 2O_2_{(g)} ------> CO_2_{(g)} +2H_2O$

4.69 2× 4.69

moles moles

$C_2H_6+ \frac{7}{2}O_2_{(g)} ------> 2CO_2_{(g)} + 3H_2O$

0.33 3.5 × 0.33

moles moles

$C_2H_4+3O_2_{(g)} ----->2CO_2+2H_2O$

0.18 3× 0.18

moles moles

Mass flow rate = 100 kg/h

Their Molar Flow rate is as follows;

$CH_4 = 4.69 k moles/h\\C_2H_6 = 0.33 k moles/h\\C_2H_4=0.18kmoles/h$

Total moles of $O_2$ required = (2 × 4.69) + (3.5 × 0.33) + (3 × 0.18) k moles

= 11.075 k moles.

In 1 mole air = 0.21 moles $O_2$

Thus, moles of air required = $\frac{1}{0.21}*11.075$

= 52.7 k mole

30% excess air = 0.3 × 52.7 k moles

= 15.81 k moles

Total air required = (52.7 + 15.81 ) k moles/h

= 68.51 k moles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

5 0

3 years ago

Which is an acceptable Lewis structure for a diatomic nitrogen molecule?

timofeeve [1]

Nitrogen has 5 valence electrons (ve-), so a diatomic nitrogen molecule will have twice as many, 10 valence electrons. Then, just draw electrons in pairs of 2 until you both get ride of all of them (reach 0) and you fill every atom (eight electrons each). It can be drawn either way, the important thing is that there are 3 electron pairs shared between the two atoms.