the half-life of a certain radioactive element is 1250 years. what percent of the atom remains after 7500 years?

The force that keeps the nucleons bound inside the nucleus of an atom

malfutka [58]

Answer:

Strong Nuclear Force

Explanation:

Strong Nuclear Force; is the force that keeps the nucleons bound inside the nucleus of an atom. It is a short range attractive force that acts between all nucleons in the nucleus of the atom that is much stronger than electrical forces.

3 0

2 years ago

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A 110.0-g sample of metal at 82.00°C is added to 110.0 g of H2O(l) at 27.00°C in an insulated container. The temperature rises t

erastovalidia [21]

Answer:

The specific heat of the metal is 0.289 J/g°C

Explanation:

Step 1: Data given

Mass of metal = 110.0 grams

Temperature of the metal = 82.00 °C

MAss of water = 110.0 grams

Temperature of the water = 27.00 °C

The final temperature = 30.56 °C

The specific heat of H2O is 4.18 J/(g°C).

Step 2: Calculate the specific heat of the metal

Heat lost = heat gained

Qlost = - Qgained

Qmetal = - Qwater

Q =m*c*ΔT

m(metal)*c(metal)*ΔT(metal) = -m(water) * c(water) *ΔT(water)

⇒with m(metal) = the mass of the metal = 110.0 grams

⇒with c(metal) = the specific heat of the metal = TO BE DETERMINED

⇒with ΔT(metal) = the change of temperature of the metal = T2 - T1 = 30.56 - 82.00 °C= -51.44 °C

⇒with m(water) = the mass of water = 110.0 grams

⇒with c(water) = the specific heat of water = 4.18 J/g°C

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = 30.56 °C - 27.00 °C = 3.56 °C

110.0 * c(metal) * -51.44 = -110.0 * 4.18 * 3.56

c(metal) = 0.289 J/g°C

The specific heat of the metal is 0.289 J/g°C

7 0

2 years ago

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Which statements correctly describe the process of nuclear fusion?

Keith_Richards [23]

It bonds atoms together to create molecules

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2 years ago

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A sample of baking soda contains 34.48 g of sodium, 1.51 g of hydrogen, 18.02 g of carbon, and 72.00 g of oxygen.

Ivahew [28]

Total mass of the sample =126.01 g

percent by mass of Sodium=27.40%

percent by mass of Hydrogen=1.198%

percent by mass of Carbon=14.300%

percent by mass of Oxygen=57.138%

Now the

Total mass of the sample of baking soda =Mass of sodium+Mass of hydrogen+mass of carbon+mass of oxygen.

Total mass of the sample of baking soda =34.48 g +1.51 g+18.02 + 72.00 g

Total mass of the sample of baking soda =126.01 g

percent by mass of each element =mass of element / Total mass of sample x 100

Sodium =34.48 g/126.01 g x 100=27.40%

Hydrogen =1.51 g/126.01 g x 100=1.198%

Carbon =18.02g/126.01 g x 100=14.300%

Oxygen =72.00g/126.01 g x 100=57.138%

See similar question here: brainly.com/question/18863998

7 0

2 years ago

The name "penicillin" is used for several closely-related antibiotics. A 1.2177 g sample of one of these compounds is burned, pr

masya89 [10]

Answer: The empirical formula for the given organic compound is $C_7H_{11}O_2SN$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$S_vN_wC_xH_yO_z+O_2\rightarrow CO_2+H_2O+SO_2+NO$

where, 'v', 'w' 'x', 'y' and 'z' are the subscripts of sulfur, nitrogen, carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=0.2829g$

Mass of $H_2O=0.1159g$

Mass of $SO_2=0.4503g$

Mass of NO = 0.2109 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

Molar mass of sulfur dioxide = 64 g/mol

Molar mass of nitrogen monoxide = 30 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 2.1654 g of carbon dioxide, $\frac{12}{44}\times 2.1654=0.590g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.6965 g of water, $\frac{2}{18}\times 0.6965=0.077g$ of hydrogen will be contained.

For calculating the mass of sulfur:

In 64 g of sulfur dioxide, 32 g of sulfur is contained.

So, in 0.4503 g of sulfur dioxide, $\frac{32}{64}\times 0.4503=0.225g$ of sulfur will be contained.

For calculating the mass of nitrogen:

In 30 g of nitrogen monoxide, 14 g of nitrogen is contained.

So, in 0.2109 g of nitrogen monoxide, $\frac{14}{30}\times 0.2109=0.098g$ of nitrogen will be contained.

Mass of oxygen in the compound = (1.2177) - (0.590 + 0.077 + 0.225 + 0.098) = 0.2277 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.590g}{12g/mole}=0.049moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.077g}{1g/mole}=0.077moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.2277g}{16g/mole}=0.0142moles$

Moles of Sulfur = $\frac{\text{Given mass of Sulfur}}{\text{Molar mass of sulfur}}=\frac{0.225g}{32g/mole}=0.007moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{0.098g}{14g/mole}=0.007moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.007 moles.

For Carbon = $\frac{0.049}{0.007}=7$

For Hydrogen = $\frac{0.077}{0.007}=11$

For Oxygen = $\frac{0.0142}{0.007}=2.03\approx 2$

For Sulfur = $\frac{0.007}{0.007}=1$

For Nitrogen = $\frac{0.007}{0.007}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O : S : N = 7 : 11 : 2 : 1 : 1

Hence, the empirical formula for the given compound is $C_7H_{11}O_2S_1N_1=C_7H_{11}O_2SN$

4 0

2 years ago