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3 years ago

What is the activity of a 52.3 μCi sample of carbon‑14 in becquerels?

Chemistry

1 answer:

Llana [10]3 years ago

3 0

Answer:

1935100 Bq

Explanation:

Let us recall that:

If 1 μCi can be equivalent to 37000 Bq

Then; the activity of 52.3 μCi will be:

$\dfrac{37000 \ Bq}{1 \ \mu Ci}\times 52.3 \ \mu Ci \\ \\ \\ \\ \mathbf{= 1935100 \ Bq}$

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Please help!! i need the answer ASAP!

nirvana33 [79]

Answer:

23.34 %.

Explanation:

The percentage of water must be calculated as a mass percent.

We need to find the mass of water, and the total mass in one mole of the compound. For that we need to use the atomic masses of each element and take in consideration the number of atoms of each element in the formula unit.

Atomic masses of the elements:

Cd: 112.411 g/mol, N: 14.0067 g/mol, O: 15.999 g/mol, and H: 1.008 g/mol.

Mass of the formula unit:

Cd(NO₃)₂•4H₂O

mass of the formula unit = (At. mass of Cd) + 2(At. mass of N) + 10(At. mass of O) + 8(At. mass of H) = (112.411 g/mol) + 2(14.0067 g/mol) + 10(15.999 g/mol) + 8(1.008 g/mol) = 308.5 g/mol.

Mass of water in the formula unit:

mass of water = (4 × 2 × 1.008 g/mol) + (4 × 15.999 g/mol) = 72.0 g/mol.

So, the percent of water in the compound = [mass of water / mass of the formula unit] × 100 = [(72.0 g/mol)/(308.5 g/mol)] × 100 = 23.34 %

7 0

3 years ago

Calcule a variação da entalpia dessa reação ( 2 NH3 (g) ---> CO(NH2)2 (s) + H2O (L) ) a partir das seguintes equações termoqu

Nitella [24]

ΔH = +438 kJ

We have three equations:

(I) N₂ + 3H₂ → 2NH₃; ΔH = -92 kJ

(II) H₂ +½O₂ → H₂O; ΔH = -286 kJ

(III) CO(NH₂)₂ + ³/₂O₂ → CO₂ + 2H₂O + N₂; ΔH = -632 kJ

From these, we must devise the target equation:

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; ΔH = ?

_________________________________

The target equation has 2NH₃ on the left, so you reverse equation (I).

When you reverse an equation, you reverse the sign of its ΔH.

(V) 2NH₃ → N₂ + 3H₂; ΔH = +92 kJ

Equation (V) has 1N₂ on the right, and that is not in the target equation.

You need an equation with 1N₂ on the left.

Reverse Equation (III).

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; ΔH = +632 kJ

Equation (VI) has ³/₂O₂ on the right, and that is not in the target equation.

You need ³/₂O₂ on the left.

Multiply Equation (II) by three.

When you multiply an equation by three, you multiply its ΔH by three.

(VII) 3H₂ +³/₂O₂ → 3H₂O; ΔH = -286 kJ

Now, you add equations (V), (VI), and (VII), cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

_______________________________________

We get the target equation (IV):

(V) 2NH₃ → N₂ + 3H₂; ΔH = + 92 kJ

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; ΔH = +632 kJ

(VII) 3H₂ +³/₂O₂ → 3H₂O; ΔH = -286 kJ

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; ΔH = +438 kJ

7 0

2 years ago

Which of the following elements will form a positive ions? check all that apply

tamaranim1 [39]

we're the rest of the question

3 0

3 years ago

Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds:(a) |(b) Sr(c) K(d) N(e) S(f)

Pavel [41]

Answer:

(a) I⁻ (charge 1-)

(b) Sr²⁺ (charge 2+)

(d) N³⁻ (charge 3-)

(e) S²⁻ (charge 2-)

(f) In³⁺ (charge 3+)

Explanation:

To predict the charge on a monoatomic ion we need to consider the octet rule: atoms will gain, lose or share electrons to complete their valence shell with 8 electrons.

(a) |

I has 7 valence electrons so it gains 1 electron to form I⁻ (charge 1-).

(b) Sr

Sr has 2 valence electrons so it loses 2 electrons to form Sr²⁺ (charge 2+).

(c) K

K has 1 valence electron so it loses 1 electron to form K⁺ (charge 1+).

(d) N

N has 5 valence electrons so it gains 3 electrons to form N³⁻ (charge 3-).

(e) S

S has 6 valence electrons so it gains 2 electrons to form S²⁻ (charge 2-).

(f) In

In has 3 valence electrons so it loses 3 electrons to form In³⁺ (charge 3+).

3 0

3 years ago

Hey pleaseeeeeeeeeeeeeeeeeeeeeeee

charle [14.2K]

Answer: B

Explanation:

6 0

3 years ago

Read 2 more answers