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Sedbober [7]
3 years ago
13

Write symbol equations for the reactions that occur when these pairs react:

Chemistry
1 answer:
timurjin [86]3 years ago
8 0

Answer:

Explanation:

word equation:

a) Bromine + potassium iodide → potassium bromide + iodine

Chemical equation:

Br₂ + KI → KBr + I₂

Balanced chemical equation:

Br₂ + 2KI → 2KBr + I₂

B) word equation:

Iodine + potassium astatide → potassium iodide + astatine

Chemical equation:

I₂ + KAt → KI + At₂

Balanced chemical equation:

I₂ + 2KAt → 2KI + At₂

c) word equation:

chlorine + sodium iodide    →  sodium chloride + iodine

Chemical equation:

 Cl₂ + NaI → NaCl + I₂

Balanced chemical equation:

Cl₂ + 2NaI → 2NaCl + I₂

d) word equation:

fluorine + sodium bromide → sodium fluoride + bromine

Chemical equation:

 F₂ + NaBr → NaF + Br₂

Balanced chemical equation:

F₂ + 2NaBr → NaF + Br₂

e) word equation:

fluorine + potassium chloride → potassium fluoride + chlorine

Chemical equation:

 F₂ + KCl → KF + Cl₂

Balanced chemical equation:

  F₂ + 2KCl → KF + Cl₂

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1) What mass of Na2CO3 is required to make 50cc of its seminormal solution?
love history [14]

Answer:

m=1.325gNa_2CO_3

Explanation:

Hello,

In this case, by considering the given seminormal solution, we infer it is a 0.5-N solution which means that we can obtain the equivalent grams as shown below for the 55 cc (0.055 L) volume:

eq-g=0.5eq-g/L*0.050L=0.025eq-g

Next, since sodium carbonate has two sodium ions with a +1 oxidation state each, we can obtain the moles:

mol=0.025eq-gNa_2CO_3*\frac{1molNa_2CO_3}{2eq-gNa_2CO_3}\\ \\mol=0.0125molNa_2CO_3

Finally, the mass is computed by using its molar mass (106 g/mol)

m=0.0125molNa_2CO_3*\frac{106gNa_2CO_3}{1molNa_2CO_3} \\\\m=1.325gNa_2CO_3

Regards.

7 0
3 years ago
Someone please help me with this ???
Verizon [17]

Answer:

c

Explanation:

6 0
2 years ago
The total pressure in a mixture of gases is equal to the partial pressures of
baherus [9]

Answer:

"The total pressure in a mixture of gases is equal to the sum of partial pressures of each gas"

Explanation:

Dalton's law of partial pressures state that, in a mixture of gases, the total pressure is equal to the sum of the partial pressure exerted by each gas of the mixture. The equation is:

Total pressure = Partial pressure Gas 1 + Partial pressure Gas 2 + .... + Partial pressure Gas n

To complete the sentence we can say:

"The total pressure in a mixture of gases is equal to the sum of partial pressures of each gas"

5 0
2 years ago
A gas sample occupies 3.25 liters at 297.5K and 2.4 atm. Determine the temperature at which the gas will occupy 4.25 L at 1.50 a
lorasvet [3.4K]

For equal moles of  gas, temperature can be calculated from ideal gas equation as follows:

P×V=n×R×T ...... (1)

Initial volume, temperature and pressure of gas is 3.25 L, 297.5 K and 2.4 atm respectively.

2.4 atm ×3.25 L=n×R×297.5 K

Rearranging,

n\times R=0.0262 atm L/K

Similarly at final pressure and volume from equation (1),

1.5 atm ×4.25 L=n×R×T

Putting the value of n×R in above equation,

1.5 atm ×4.25 L=0.0262 (atm L/K)×T

Thus, T=243.32 K


7 0
2 years ago
What is the volume occupied by 4.20 miles of oxygen gas (O2) at STP
Drupady [299]

Answer: 94.13 L

Explanation: In STP in an ideal gas there is a standard value for both temperature and pressure. At STP,pressure is equal to 1atm and the temperature at 0°C is equal to 273.15K. This problem is an ideal gas so we use PV=nRT where R is a constant R= 0.08205 L.atm/mol.K.

To find volume, derive the equation, it becomes V=nRT/P. Substitute the values. V= 4.20 mol( 0.08205L.atm/mol.K)(273.15K) / 1 atm = 94.13 L. The mole units, atm and K will be cancelled out and L will be the remaining unit which is for volume.

8 0
2 years ago
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