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lawyer [7]
4 years ago
10

When 1 molecule of calcium phosphate is added to water, what and how many ions are formed? remember that ions are charged and so

you must indicate the correct charge on each ion to receive credit for this question?

Chemistry
1 answer:
Fiesta28 [93]4 years ago
3 0

Asnwer :When 1 molecule of calcium phosphate is added to water, total 5 ions are formed 3 ions of Calcium(Ca^{2+}) and 2 ions of Phosphate(PO_{4}^{3-})

Explanation : When Calcium phosphate is added to water it gets dissociated into ions. Calcium phosphate (Ca_{3}(PO_{4})_{2}) dissociates into 3 ions of Calcium Ca^{2+} and 2 ions of Phosphate PO_{4}^{3-}.

Dissociation equation of calcium phosphate is :

1 Ca_{3}(PO_{4})_{2} \leftrightarrow 3 Ca^{2+} + 2PO_{4}^{3-}

From the dissociation equation we can conclude that total 5 ions are formed when 1 molecule of calcium phosphate is added to water.

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How molecules of N2 gas can be present in a 2.5 L flask at 50°C and 650 mmHg?
ratelena [41]

Answer:

0.482 ×10²³ molecules

Explanation:

Given data:

Volume of gas = 2.5 L

Temperature of gas = 50°C (50+273 = 323 k)

Pressure of gas = 650 mmHg (650/760 =0.86 atm)

Molecules of N₂= ?

Solution:

PV= nRT

n = PV/RT

n = 0.86 atm × 2.5 L /0.0821 atm. mol⁻¹. k⁻¹. L × 323 k

n = 2.15 atm. L /26.52 atm. mol⁻¹.L

n = 0.08 mol

Number of moles of N₂ are 0.08 mol.

Number of molecules:

one mole = 6.022 ×10²³ molecules

0.08×6.022 ×10²³ = 0.482 ×10²³ molecules

5 0
3 years ago
Which forms when a humid air mass rises into a cooler temperature area?
Zielflug [23.3K]

Option A: Clouds

In the morning, air is cool and as sun begins to rise it starts increasing the temperature of air. By time, the air becomes warmer and warmer. Depending upon the surrounding conditions, air in different areas heat up at different rates.

Due to this heating, thermal formation takes place, this is due to uneven heating of surface of earth. The thermal formation at surface causes difference in temperature of surface of the earth and air around it. The warm air has tendency to rise thus, the air in the thermal rise and expand. Due to expansion it cools down, this process continues till the temperature of thermal air reaches equals to the temperature of surrounding air. This results in the formation of cloud.

Thus, when a humid air mass rises into a cooler temperature area, clouds formation takes place



3 0
3 years ago
Read 2 more answers
You hold two magnets together, end to end, and then let go. If the magnets stay together, the magnets are being _____.
Aleksandr-060686 [28]

Answer:

C-attracted.

Explanation:

When two magnets stay together they are being attracted. If they are pushing each other away, they are being repelled.

5 0
3 years ago
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If you have 177 grams of C4H80, how many moles do you have?
quester [9]
I think it’s 2.45, not 100% sure
4 0
3 years ago
The Balmer series, named after Johann Balmer, is a portion of the hydrogen emission spectrum produced from the transitions betwe
horrorfan [7]

Explanation:

The wavelength of the balmer series is calculated using the following steps;

- Find the Principle Quantum Number for the Transition

- Calculate the Term in Brackets

- Multiply by the Rydberg Constant

- Find the Wavelength

The Balmer series in a hydrogen atom relates the possible electron transitions down to the n = 2 position to the wavelength of the emission that scientists observe.

The λ symbol represents the wavelength, and RH is the Rydberg constant for hydrogen, with RH = 1.0968 × 107 m−1

n=7 to n=2

- The principal quantum numbers are 2 and 7.

-  (1/2²) − (1 / n²₂)

For n₂ = 7, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 7²)

= (1/4) − (1/49)

= 0.2230

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.2230

= 2445864 m−1

- λ = 1 / 2445864 m−1

= 4.08 × 10−7 m

= 408 nanometers

≈ 410nm

n=6 to n=2

- The principal quantum numbers are 2 and 6.

-  (1/2²) − (1 / n²₂)

For n₂ = 6, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 6²)

= (1/4) − (1/36)

=  0.2222

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 3/16

= 2437090 m−1

- λ = 1 / 2437090 m−1

= 4.10 × 10−7 m

= 410 nanometers

n=5 to n=2

- The principal quantum numbers are 2 and 5.

-  (1/2²) − (1 / n²₂)

For n₂ = 5, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 5²)

= (1/4) − (1/25)

= 0.21

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.21

= 2303280 m−1

- λ = 1 / 2303280 m−1

= 4.34 × 10−7 m

= 434 nanometers

n=4 to n=2

- The principal quantum numbers are 2 and 4.

-  (1/2²) − (1 / n²₂)

For n₂ = 4, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 4²)

= (1/4) − (1/16)

= 0.1875

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.1875

= 2056500 m−1

- λ = 1 / 2056500 m−1

= 4.86 × 10−7 m

= 486 nanometers

n=3 to n=2

- The principal quantum numbers are 2 and 3.

-  (1/2²) − (1 / n²₂)

For n₂ = 3, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 3²)

= (1/4) − (1/9)

= 0.13889

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.13889

= 1523345 m−1

- λ = 1 / 1523345 m−1

= 6.56 × 10−7 m

= 656 nanometers

7 0
3 years ago
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