How many moles of oxygen gas are required to form 55g of carbon monoxide gas

The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be

Stels [109]

Explanation:

1) Initial mass of the Cesium-137= $N_o$ = 180 mg

Mass of Cesium after time t = N

Formula used :

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

Half life of the cesium-137 = $t_{1/2}=30 years[p/tex]where,
[tex]N_o$ = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

$t_{\frac{1}{2}}$ = half life of the isotope

$\lambda$ = rate constant

$N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}$

Now put all the given values in this formula, we get

$N=180mg\times e^{-\frac{0.693}{30 years}\times t}$

Mass that remains after t years.

$N=180 mg\times e^{0.0231 year^{-1}\times t}$

Therefore, the parent isotope remain after one half life will be, 100 grams.

2)

t = 70 years

$N_o=180 mg$

$t_{1/2}= 30 yeras$

$N=180mg\times e^{-\frac{0.693}{30 years}\times 70 years}$

N = 35.73 mg

35.73 mg of cesium-137 will remain after 70 years.

3)

$N_o=180 mg$

$t_{1/2}= 30 yeras$

N = 1 mg

t = ?

$1 mg =180mg\times e^{-\frac{0.693}{30 years}\times t}$

$\frac{-30 year}{0.693}\times \ln \frac{1 mg}{180 mg}=t$

t = 224.80 years ≈ 225 years

After 225 years only 1 mg of cesium-137 will remain.

7 0

3 years ago

Al2(SO4)3+Mg(NO3)2⟶ What would be the product(s) of this reaction? *These are NOT balanced, just look for the correct products*

Readme [11.4K]

Answer:

$Al_2(SO_4)_3+3Mg(NO_3)_2\rightarrow 2Al(NO_3)_3+3MgSO_4$

Explanation:

Hello there!

In this case, for the given reactants side, we infer this is a double replacement reaction because all the cations and anions are switched around as a result of the chemical change, we infer that the products side include aluminum with nitrate and magnesium with sulfate as shown below:

$Al_2(SO_4)_3+Mg(NO_3)_2\rightarrow Al(NO_3)_3+MgSO_4$

However, we need to balance since unequal number of atoms are present at both sides, thus, we do that as shown below:

$Al_2(SO_4)_3+3Mg(NO_3)_2\rightarrow 2Al(NO_3)_3+3MgSO_4$

Thus, we make 6 Al atoms, 3 S atoms, 3 Mg atoms and 30 O atoms on each side in agreement with the law of conservation of mass.

Regards!

7 0

2 years ago

A sample compound contains 9.11 g Ni and 5.89g F. What is the empirical formula of this compound?

Alexxx [7]

The empirical formula of the compound is C. NiF₂.

Step 1. Calculate the moles of each element

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our job is to calculate the molar ratio of Ni to F.

Moles of Ni = 9.11 g Ni × (1 mol Ni /(58.69 g Ni) = 0.1552 mol Ni

Moles of F = 5.89 g F × (1 mol F/19.00 g F) = 0.3100 mol F

Step 2. Calculate the molar ratio of the elements

Divide each number by the smallest number of moles

Ni:F = 0.1552:0.3100 = 1:1.997 ≈ 1:2

Step 3: Write the empirical formula

EF = NiF₂

8 0

3 years ago

Read 2 more answers

Benzene has a specific gravity of 0.88. if it was spilled in a river what would it do?

Gelneren [198K]

As we know,
Density of Benzene = 876 Kg/m³
And,
Density of Water = 997 Kg/m³
So,
Specific Gravity is calculated as,

Specific Gravity = Density of Benzene / Density of Water

Specific Gravity = 876 Kg/m³ / 997 Kg/m³

Specific Gravity = 0.878

Every object having specific gravity less than 1 will float on water and if value is greater than 1 then it will sink.

Benzene being non-polar in nature does not mix with water and due to less density it will float on the surface of water.

4 0

3 years ago

How much energy is released by the decay of 3 grams of 230Th in the following reaction 230 Th - 226Ra + "He (230 Th = 229.9837 g

Serhud [2]

Answer: The energy released for the decay of 3 grams of 230-Thorium is $2.728\times 10^{-15}J$