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3 years ago

Please answer asap for brainliest!!

1 answer:

4 0

The reactant in a chemical reaction that remains when a reaction stops when the limiting reactant is fully consumed.Percent yield is the percent ratio of actual yield to theoretical yield. Experimental yield divides by the theoretical yield and it multiplies by 100%

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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:

Step 1: 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

Step 2: 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

I need help with this question ASAP

sladkih [1.3K]

the answer is thermal.......

4 0

2 years ago

So its not A so its B C or D plz help

Rainbow [258]

Answer:

the answer is B

hope this helped!

3 0

3 years ago

A cube of salt is crushed before being stirred into water.

andrew-mc [135]

Physical change

Giddy Up!!!!!

7 0

3 years ago

Read 2 more answers

For the following electrochemical reaction: Al3+(aq) + 3e -> Al(s) Eº = -1.66 V E° = 2.87 F2(g) + 2e -> 2F (aq) Calculate

makvit [3.9K]

Answer: The standard electrode potential of the cell is 4.53 V.

Explanation:

We are given:

$E^o_{(F_2/F^-)}=2.87V\\E^o_{(Al^{3+}/Al)}=-1.66V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Aluminium will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=2.87-(-1.66)=4.53V$

Hence, the standard electrode potential of the cell is 4.53 V.

6 0

2 years ago