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irakobra [83]
3 years ago
14

Can someone please answer my last question, ill post more like it for more points :)

Chemistry
1 answer:
fiasKO [112]3 years ago
6 0
I will give me a second:)))))
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A square chunk of plastic has a length of 5 cm, width of 5 cm and height of 5 cm. It has a mass of 200 g. What is it’s density
vichka [17]

Answer:

1.6g/mL

Explanation:

Density equation is D=m/v

Density = g/mL

m=mass of sample in grams

v = volume of sample in mL

The volume of a square can be calculated by V=l*w*h.

In this case it is 5cm*5cm*5cm = 125cm^3

Since we know that 1cm^3 ~ 1mL we can convert the volume to mL as so:

125cm^3 (1mL/(1cm^3)) = 125mL

Then simply plug into the density equation:

D=200g/125mL = 1.6g/mL

3 0
3 years ago
Which statement about scientific notation is true?
vladimir2022 [97]

Explanation:

the proper format for scientific notation the following conditions must be true

8 0
3 years ago
The best tool to use to locate the boundaries between countries is a(n) _____.
8_murik_8 [283]

The correct answer is (b) political map

The explanation:

when a political map: is different than other types of maps because it focuses on government or administrative boundaries rather than geographical or physical features. Instead of showing viewers what exists in the land, it shows those imaginary lines that serve to separate countries, states, territories, and cities. These maps generally include larger bodies of water, such as oceans, rivers, and lakes, as landmarks.

It was made by man not natural like the others maps.

so this is the correct answer.

8 0
3 years ago
Manganese nodules are considered a valuable resource from the ocean floor because they are abundant and contain not only mangane
MAVERICK [17]

Answer: iron (oxides of iron)

Explanation: manganese nodules usually contain layers of iron and manganese oxides in a concentric arrangement and are called polymetallic nodules.

4 0
3 years ago
Use the given data at 500 K to calculate ΔG°for the reaction
Anton [14]

Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

3 0
3 years ago
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