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3 years ago

Substance P is carbon.

Chemistry

1 answer:

stepan [7]3 years ago

4 0

Explanation:

substance Q could be oxygen (O2)

substance R could be carbon dioxide (CO2)

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A 50.0 mL graduated cylinder has a mass of 65.1 g. When it is filled with an unknown liquid to the 49.3 mL mark, the cylinder an

Delvig [45]

Answer: The density of liquid is $1.12g/cm^3$

Explanation:

We are given:

Mass of cylinder, $m_1$ = 65.1 g

Mass of liquid and cylinder combined, M = 120.5 g

Mass of liquid, $m_2$ = $(M-m_1)=(120.5-65.1)g=55.4g$

To calculate density of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

We are given:

Mass of liquid = 55.4 g

Volume of liquid = 49.3 mL = $49.3 cm^3$ (Conversion factor: $1mL=1cm^3$ )

Putting values in above equation, we get:

$\text{Density of liquid}=\frac{55.4g}{49.3cm^3}\\\\\text{Density of liquid}=1.12g/cm^3$

Hence, the density of liquid is $1.12g/cm^3$

3 0

3 years ago

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti

gayaneshka [121]

Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

$CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)$

Now we have to calculate value of $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

where,

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(HCH_3CO_2(g))}$ = -389.8 kJ/mol

$\Delta G^0_{(CH_3OH(g))}$ = -161.96 kJ/mol

$\Delta G^0_{(CO(g))}$ = -137.2 kJ/mol

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

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

$K$ = equilibrium constant = ?

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

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

4 0

3 years ago

I really need help with this. Be serious. If you answer something random just for the points I will report you to brainly.

sashaice [31]

Answer:

It get thicker beacause the mid ocean

Explanation:

Because when it gets moved back the heat rises and it builds up to be thicker.

8 0

2 years ago

What is the application of chemistry

vlabodo [156]

Answer:

Chemistry plays an important and useful role towards the development and growth of a number of industries. This includes industries like glass, cement, paper, textile, leather, dye etc. We also see huge applications of chemistry in industries like paints, pigments, petroleum, sugar, plastics, Pharmaceuticals.

3 0

3 years ago

Balance the following reaction:

kvv77 [185]

C_5H_8+13/2O_2—»5CO_2+4H_2O

Balanced one

2C_5H_8+13O_2—»10CO_2+8H_2O

Moles of Pentyne

Given mass/Molarmass
34/68
0.5mol

Moles of H_2O

8/2(0.5)
4(0.5)
2mol

1mol releases 241.8KJ

2mol releases 241.8(2)=483.6KJ

8 0

2 years ago