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Dimas [21]
3 years ago
9

What product(s) are formed during this chemical reaction? (Will be two answers)

Chemistry
1 answer:
lianna [129]3 years ago
3 0

Answer:

BK. BK. BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK BK

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Help anyone can help me do this question,I will mark brainlest.​
Anettt [7]

Answer:

12 grams

Explanation:

if  you used the transformation equation  you would end up with 12 grams

4 0
3 years ago
For the reaction, 2Cr2+ + Cl2(g) ---> 2Cr3+ + 2Cl- Ecell (standard conditions) = 1.78V
Lapatulllka [165]

Answer:

Eºcell = -1.78 V

Explanation:

The Eº cell is an intensive property, i.e they do not depend on the quantity of material present and the desired reaction in our problem  is exactly half the reverse of the one given, the Eºcell will then  be the negative of the first then Eºcell is -1.78 V and the redox reaction will be non-spontaneous as opposed to the first.

5 0
3 years ago
What are the spectator ions in the
bonufazy [111]

Answer:

S

Explanation:

5 0
2 years ago
7.Calculate the volume of nitrogen that reacts with 12dm3 of hydrogen with the volume of both gases measured at rtp:
marta [7]

Answer:

4.03dm³

Explanation:

The reaction expression is given as:

       3H₂   +   N₂   →    2NH₃  

  Volume of hydrogen  = 12dm³  

AT rtp:

             1 mole of gas occupies volume of 22.4dm³  

             x mole of hydrogen will occupy a volume of 12dm³

     Number of moles of hydrogen  = \frac{12}{22.4}   = 0.54mole

From the balanced reaction equation:

            3 mole of hydrogen gas combines with 1 mole of Nitrogen gas

         0.54 mole of hydrogen as will therefore combine with \frac{0.54}{3}   = 0.18moles of nitrogen gas

Since ;

                     1 mole of gas occupies a volume of 22.4dm³

               0.18moles of Nitrogen gas will occupy 0.18 x 22.4  = 4.03dm³

6 0
3 years ago
A volume of 80.0 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. I
murzikaleks [220]

Answer:

The mass of the steel bar is 26.833 grams

Explanation:

<u>Step 1: </u>data given

ΣQ gained = ΣQ lost

Q=m*C*ΔT

with m = mass in grams

with C= specific heat capacity ( in J/(g°C))

with ΔT = change in temperature = T2-T1

Qsteel = Qwater

msteel * Csteel * (T2steel - T1steel) = mwater * Cwater * (T2water - T1water)

Mass of steel = TO BE  DETERMINED

mass of water =⇒ since 1mL = 1g : 80 mL = 80g

Csteel =0.452 J/(g °C

Cwater = 4.18 J/(g °C

initial temperature steel T1 : 2 °C

final temperature steel T2 = 21.3 °C

initial temperature water T1 =22 °C

final temperature water T2 = 21.3 °C

<u>Step 2:</u> Calculate mass of steel

msteel * Csteel * (T2steel - T1steel) = mwater * Cwater * (T2water - T1water)

msteel * 0.452 *(21.3-2) = 80 * 4.18 * (21.3-22)

msteel = (80 * 4.18 * (-0.7)) / (0.452 * 19.3)

msteel = -234.08 / 8.7236

msteel = -26.833 g

Since mass can't be negative we should take the absolute value of it = 26.833g

The mass of the steel bar is 26.833 grams

6 0
3 years ago
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