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3 years ago

Cobalt III hydroxide and nitric acid react according to the following balanced equation:

1 answer:

6 0

The mass of cobalt (III) needed is
m = 5.2 L (0.42 mol/L) ( 93 g/mol)
m = 97.65 g

The volume of nitric acid needed is
V = 5.2 L (0.42 mol/L) (3 mol / 1 mol) (1000 mL/1.6 mol)
V = 1968.75 mL

The moles of water produced is
n = 5.2 L (0.42 mol/L) (3 mol / 1 mol)
n = 3.15 moles

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3 years ago

How much heat energy is needed to heat 300g of water from 10 degrees Celsius to 50 degrees Celsius

elixir [45]

Answer:

There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C

Explanation:

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

mass of water = 300 grams

initial temperature = 10°C

final temperature = 50°C

Temperature rise = 50 °C - 10 °C = 40 °C

Specific heat capacity of water = 4.184 J/g °C

<u>Step 2:</u> Calculate the heat

Q = m*c*ΔT

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There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C

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3 years ago

Two objects are brought into contact Object 1 has mass 0.76 kg, specific heat capacity 0.87) g'c and initial temperature 52.2 'C

taurus [48]

Answer:

$T_F=77.4\°C$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:

$Q_1=-Q_2$

In terms of mass, specific heat and temperature change is:

$m_1C_1(T_F-T_1)=-m_2C_2(T_F-T_2)$

Now, solve for the final temperature, as follows:

$T_F=\frac{m_1C_1T_1+m_2C_2T_2}{m_1C_1+m_2C_2}$

Then, plug in the masses, specific heat and temperatures to obtain:

$T_F=\frac{760g*0.87\frac{J}{g\°C} *52.2\°C+70.7g*3.071\frac{J}{g\°C}*154\°C}{760g*0.87\frac{J}{g\°C} +70.7g*3.071\frac{J}{g\°C}} \\\\T_F=77.4\°C$

Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.

Regards!

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3 years ago

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telo118 [61]

Answer:

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Explanation:

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