Which reaction is an example of an endothermic reaction?

How many moles are in 2.8 x 1023 BS3 molecules?

stiks02 [169]

Answer:

The atomic mass given on a periodic table that is given in grams is the mass of one mole (6.022 × 1023 particles) of that element. EXAMPLE: As you can see from the example above, one mole of Carbon would have a mass of 12.011 grams.

6 0

3 years ago

A compound contains only carbon, hydrogen, and oxygen. combustion of 11.75 mg of the compound yields 17.61 mg co2 and 4.81 mg h2

Ymorist [56]

Number of moles is defined as the ratio of given mass in g to the molar mass.

First, convert the given mass of carbon dioxide in mg to g:

1 mg = 0.001 g

17.61 mg = 0.01761 g

Number of moles of carbon dioxide = $\frac{0.01761 g}{44.01 g/mol}$

= $0.0004001 mol$

Mass of carbon = number of moles of carbon dioxide \times molar mass of carbon

= $0.0004001 mol\times 12.011 g/mol$

= $0.004806 g$

Number of moles of water= $\frac{0.00481 g}{18 g/mol}$

= $2.672\times 10^{-4}$

Since, water contains two hydrogen atoms. Thus,

Moles of hydrogen = $2\times 2.672\times 10^{-4}$

= $5.34\times 10^{-4}$

Mass of hydrogen = $5.34\times 10^{-4}\times \times 1.008 g/mol$

= $5.34\times 10^{-4} g$

Mass of oxygen = $0.001175-(5.38\times 10^{-4}g+0.004806 g)$

= $0.006405 g$

Number of moles of oxygen = $\frac{0.006405 g}{15.999 g/mol}$

= $0.000400$

Now,

$C_{0.0004001} H_{0.000534} O_{0.000400}$

Divide the smallest number to get the whole number,

$C_{\frac{0.0004001}{0.000400}} H_{\frac{0.000534}{0.000400}} O_{\frac{0.000400}{0.000400}}$

we get,

$C_{1} H_{1.33} O_{1}$

Now, multiply all the subscript by 3 to get the whole number,

$C_{3} H_{4} O_{3}$ (empirical fomula)

Molar mass of the compound = $3\times 12.011 g/mol+4\times 1.008 g/mol+3\times 15.999 g/mol$

= $88.062 g/mol$

Divide given molar mass of the compound with the molar mass of the compound.

= $\frac{176.1 g/mol}{88.062 g/mol}$

= $1.999\simeq 2$

Thus, multiply the subscripts of empirical formula by 2 to get the molecular formula, we get:

$C_{6}H_{8}O_{6}$

Hence, empirical formula is $C_{3}H_{4}O_{3}$ and molecular formula is $C_{6}H_{8}O_{6}$

8 0

3 years ago

How can you tell states of matter apart?

sergejj [24]

You can tell them apart from its form.

If it forms loose and needs like a container to hold it up, that's liquid.

If it is hard and stable, definitely a solid.

And if you barely see something very loose (and normally goes up), that's liquid.

8 0

3 years ago

Read 2 more answers

A certain element exists as two different isotopes. 90.50% of its atoms have a mass of 20.00 amu and 9.500% of its atoms have a

Serga [27]

Answer:

20.19 amu.

Explanation:

From the question given above, the following data were obtained:

Isotope A:

Abundance (A%) = 90.5%

Mass of A = 20 amu

Isotope B:

Abundance (B%) = 9.5%

Mass of B = 22 amu

Average atomic mass =..?

The average atomic mass of the element can be obtained as follow:

Average atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]

Average atomic mass = [(20 × 90.5)/100] + [(22 × 9.5)/100]

Average atomic mass = 18.1 + 2.09

Average atomic mass = 20.19 amu

Therefore, the average atomic mass of the element is 20.19 amu

6 0

3 years ago

You have 47.0 mL of a 0.400 M stock solution that must be diluted to 0.100 M. Assuming the volumes are additive, how much water

Bingel [31]

Hey there!

Moles of stock solution:

47.0 mL in liters : 47.0 / 1000 => 0.047 L

n = M * V

n = 0.400 * 0.047

n = 0.0188 moles

Volume final :

M = n / V

0.100 = 0.0188 / V

V = 0.0188 / 0.100

V = 0.188 L => 188 mL

Therefore Watter added :

volume final - volume initial

188 mL - 47.0 mL => 141 mL

Hope That helps!

5 0

3 years ago