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3 years ago

The following atoms are all variations of the element carbon: 12C, 13C, and 14C due to differing

1 answer:

5 0

Isotopes.
Some isotopes of elements that are not usually radioactive are radioactive because they don't have a suitable number of neutrons. For example carbon-12 is stable but carbon-14 is radioactive. Carbon-12 is much more common.

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A piece of unknown metal with mass 30 g is heated to 110.0 °C and dropped into 100.0 g of water at 20.0 °C. The final temperatur

Ymorist [56]

Answer: The specific heat of metal is 0.821 J/g°C

Explanation:

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of metal = 30 g

$m_2$ = mass of water = 100 g

$T_{final}$ = final temperature = 25°C

$T_1$ = initial temperature of metal = 110°C

$T_2$ = initial temperature of water = 20.0°C

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$30\times c_1\times (25-110)=-[100\times 4.186\times (25-20)]$

$c_1=0.821J/g^oC$

Hence, the specific heat of metal is 0.821 J/g°C

8 0

3 years ago

In a space ship circling the moon , with oxygen for humans breathing. All of a sudden the ship has no oxygen can a person get en

Papessa [141]

Answer:

Explanation:

6 0

3 years ago

How many grams of lead (ll) chloride are produced from the reaction of 25g of sodium chloride according to the balanced reaction

Natalka [10]

Answer:

The answer to your question is: 25 g of PbCl2

Explanation:

Data

NaCl = 25 g

PbCl₂ = ?

Pb(NO3)2 + 2 NaCl ⇒ PbCl2 + NaNO3

MW NaCl = 58.5 g

MW PbCl2 = 277 g

2(58.5 g) of NaCl ------------------------ 277 g of PbCl2

25 g of NaCl ------------------------ x

x = (25 x 277) / 117

x = 25 g

7 0

3 years ago

An experiment requires 54.6 g of ethylene glycol, a liquid whose density in SI units is 1114 kg/m3. Rather than weigh the sample

laila [671]

Answer: 49.0 cm^3

Explanation:

1) formula: density = mass / volume

=> volume = mass / density

2) mass (given) = 54.6 g

3) density (given) = 1114 kg / m^3

conversion of units: 1114 kg/ m^3 * 1000 g/kg * 1 m^3 / (1,000,000 cm^3) = 1.114 g/cm^3

4) calculation:

volume = 54.6 g / 1.114 g/cm^3 = 49.0 cm^3

5 0

3 years ago

Read 2 more answers

Determine the quantum numbers n and l of the electron removed when bi is ionized to bi+

lozanna [386]

Electronic configuration of bismuth is given as:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^{6}4f^{14}5d^{10}6s^{2}6p^{3}$

When electron is removed from bismuth then bismuth is ionized to $Bi^{+}$ , the electronic configuration becomes:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^{6}4f^{14}5d^{10}6s^{2}6p^{2}$

Now, n represents the principal quantum number which determines the distance from the orbital of the electrons i.e. size of the orbital and its energy.

Thus, outer most electron is present in shell number 6. Thus, principal quantum number is (n=6).

Now, l represents the azimuthual quantum number which determines the shape of an orbital.

Now, if the outer most electron is present in s orbital then the l =0, if the outer most electron is present in p orbital then the l =1, if the outer most electron is present in d orbital then the l =2, if the outer most electron is present in f orbital then the l =4 and if the outer most electron is present in d orbital then the l =2, if the outer most electron is present in h orbital then the l =5.

l 0 1 2 3 4 5

Letter s p d f h

Thus, azimuthual quantum number is 1 as the outer most electron is present in the p orbital.

5 0

3 years ago