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3 years ago

List the steps in the process of calculating average atomic mass given data about the isotopes of an element.

1 answer:

Nutka1998 [239]3 years ago

5 0

To calculate the atomic mass of an element, the atomic masses of their isotopes must be considered. Their abundances must also be taken into account. This is the formula used for the calculation:
avg. atomic mass= ∑ (isotope × abundance).

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Calculate the concentration of OH-in a solution that has a concentration of H+ = 8.1 x 10^−6 M at 25°C. Multiply the answer you

Nat2105 [25]

Answer:

The answer is 12.35

Explanation:

From the question we are given that the concentration of $H^{+}$ is $8.1 * 18^{-6}M$

Generally The rate equation is given as

$K_{w} = [H^{+} ][OH^{-} ]$

and $K_{w}$ the rate constant has a value $1 * 10^{-14}$

Substituting and making [ $OH^{-}$ ] the subject we have

$[OH^{-} ] = \frac{1 * 10^{-14}}{[H^{+}]} = \frac{1 * 10^{-14}}{8.1 *10^{-6}} =1.235 * 10^{-9}$

$[OH ^ {-}] = 1.235 * 10^{-9}M$

Multiply the value by $10^{10}$ as instructed from the question we have

Answer = $1.235 * 10 ^{-9} * 10^{10} = 12.35$

Hence the answer in 2 decimal places is 12.35

7 0

4 years ago

Phosphoric acid, which is commonly used as rust inhibitor, food additive and etching agent for dental and orthopedic use, can be

Sphinxa [80]

Answer:

$P_4_{(I)}+5O_2_{(g)}+6H_2O_{(l)}\rightarrow +4H_3PO_4_{(l)}$

Explanation:

The first step is:

$P_4_{(I)}+5O_2_{(g)}\rightarrow 2P_2O_5_{(g)}$

Second step is:

$P_2O_5_{(g)}+3H_2O_{(l)}\rightarrow 2H_3PO_4_{(l)}$

Multiplying second step by 2, and adding both the steps, we get that:

$P_4_{(I)}+5O_2_{(g)}+2P_2O_5_{(g)}+6H_2O_{(l)}\rightarrow 2P_2O_5_{(g)}+4H_3PO_4_{(l)}$

Cancelling common species, we get that:

$P_4_{(l)}+5O_2_{(g)}+6H_2O_{(l)}\rightarrow +4H_3PO_4_{(l)}$

6 0

4 years ago

A 48.3 mL sample of gas in a cylinder is warmed from 22 °C to

goldenfox [79]

Answer:

58.94 mL

Explanation:

V1 = 48.3 mL V2 = v mL

T1 = 22 degree celsius OR 295 k T2 = 87 degree celsius OR 360 k

We will use the gas equation:

PV = nRT

Since the Pressure (p) , number of moles (n) and the universal gas constant(R) are all constants in this given scenario,

we can say that

V / T = k , (where k is a constant)

Since this is the first case,

V1 / T1 = k --------------------(1)

For case 2:

Since we have the same constants, the equation will be the same

V / T = k (where k is the same constant from before)

V2 / T2 = k (Since this is the second case) ------------------(2)

From (1) and (2):

V1 / T1 = V2 / T2

Now, replacing the variables with the given values

48.3 / 295 = v / 360

v = 48.3*360 / 295

v = 58.94 mL

Therefore, the final volume of the gas is 58.94 mL

4 0

3 years ago

a sample of ammonia liberates 5.66 kj of heat as it solidifies at its melting point . what is the mass of the sample?

Reil [10]

The correct answer for the question that is being presented above is this one: "16.728 g."

Given that
ΔHsolid = -5.66 kJ/mol.
This means that 5.66 kJ of heat is released when 1 mole of NH3 solidifies

When 5.57 kJ of heat is released
amount of NH3 solidifies = 5.57/5.66 = 0.984 moles

molar mass of NH3 = 17 g/mole 
1 mole of NH3 = 17 g 
So, 0.984 moles of NH3 = 17 X 0.984 = 16.728 g

7 0

3 years ago

The radioisotope that has the longest half-life is the best to use in powering planet and space exploration vehicles because the

Diano4ka-milaya [45]

Answer:

U-238

Explanation:

For a given radioisotope, half life can be defined as the time taken for the isotope to decay into one-half of its original amount. Mathematically, this is expressed as:

$t_{1/2} = \frac{0.693}{k}$

where k = rate constant for the radioactive decay process

Greater the t1/2 longer will be its stability.

Based on the given data, U-238 has the largest half life and therefore will be best suited for applications mentioned.

4 0

4 years ago

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