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Naddika [18.5K]
3 years ago
11

Please Help!

Chemistry
2 answers:
Angelina_Jolie [31]3 years ago
6 0

Answer:

Cd(NO₃)₂ + 2NH₄Cl → CdCl₂ + 2NH₄NO₃

Explanation:

The reaction given is a double replacement reaction, which means that the reactants will dissociate and the ions will be changed between them. So, let's identify the ions.

At Cd(NO₃)₂, the metal Cd has cation Cd⁺², and the anion is NO₃⁻ (the number of ions is changed and the cation is the first to appear in the formula). At NH₄Cl, the cation is NH₄⁺ and the anion is Cl⁻, so the products are CdCl₂ and NH₄NO₃ (the cation comes first, and the charges are replaced between them).

So, the reacion will be:

Cd(NO₃)₂ + NH₄Cl → CdCl₂ + NH₄NO₃

To balance the equation, all the elements must have the same number on both sides, thus we need to multiply NH₄Cl and NH₄NO₃ by 2:

Cd(NO₃)₂ + 2NH₄Cl → CdCl₂ + 2NH₄NO₃

Zarrin [17]3 years ago
5 0

CdCl2+2NH4NO3

It is substitution reaction

Cd is metal with 2+ which connect with Cl(-)

NO3 with NH4 form salt

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A sample of carbon dioxide at RTP is 0.50 dm3. How many grams of carbon dioxide do we have?
prohojiy [21]

Answer:

0.924 g

Explanation:

The following data were obtained from the question:

Volume of CO2 at RTP = 0.50 dm³

Mass of CO2 =?

Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:

1 mole of gas = 24 dm³ at RTP

Thus,

1 mole of CO2 occupies 24 dm³ at RTP.

Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e

Xmol of CO2 = 0.5 /24

Xmol of CO2 = 0.021 mole

Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.

Finally, we shall determine the mass of CO2 as follow:

Mole of CO2 = 0.021 mole

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0.021 = mass of CO2 /44

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Answer:

1. 136 °C.

2. 0.21 atm.

Explanation:

1. Determination of the new temperature in °C.

Initial volume (V1) = 1.35L

Final volume (V2) = 1.95L

Initial temperature (T1) = 283 K

Final temperature (T2) =...?

Using the Charles' law equation, the new temperature of the gas can be obtained as follow:

V1 /T1 = V2 /T2

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1.35 × T2 = 283 × 1.95

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T (°C) = T(K) – 273

T(K) = 409 K

T (°C) = 409 – 273

T (°C) = 136 °C

Therefore, the new temperature of the gas is 136 °C.

2. Determination of the new pressure.

Initial pressure (P1) = 1.34 atm

Initial volume (V1) = 267 mL

Final volume (V2) = 1.67 L

Final pressure (P2) =.?

Next, we shall convert 1.67 L to millilitres (mL). This can be obtained as follow:

1 L = 1000 mL

Therefore,

1.67 L = 1.67 L × 1000 mL / 1 L

1.67 L = 1670 mL

Therefore, 1.67 L is equivalent to 1670 mL.

Finally, we shall determine the new pressure of the gas as follow:

Initial pressure (P1) = 1.34 atm

Initial volume (V1) = 267 mL

Final volume (V2) = 1670 mL

Final pressure (P2) =.?

P1V1 = P2V2

1.34 × 267 = P2 × 1670

357.78 = P2 × 1670

Divide both side by 1670.

P2 = 357.78 / 1670

P2 = 0.21 atm.

Therefore, the new pressure of the gas is 0.21 atm.

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