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3 years ago

14

Is bioluminescence an endothermic process bioluminescence an endothermic process or exothermic process?\?

1 answer:

vichka [17]3 years ago

6 0

Bioluminescence is a chemical reaction that takes place in a living organism when the organism emits light. Bioluminescence doesn't consume or produce heat, hence it can be neither endothermic nor exothermic reaction. It emits a heatless light not heat.
For bioluminescence, it could be said that it is an exoenergetic reaction because it uses energy from a chemical reaction to produce light.

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A 275 g sample of a metal requires 10.75 kJ to change its temperature from 21.2 oC to its melting temperature, 327.5 oC. What is

Vlada [557]

Answer:

$c=0.127\ J/g^{\circ} C$

Explanation:

Given that,

Mass of the sample, m = 275 g

It required 10.75 kJ of heat to change its temperature from 21.2 °C to its melting temperature, 327.5 °C.

We need to find the specific heat of the metal. The heat required by a metal sample is given by :

$Q=mc\Delta T$

c is specific heat of the metal

$c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{10.75\times 10^3\ J}{275\times (327.5 -21.2)}\\\\=0.127\ J/g^{\circ} C$

So, the specific heat of metal is $0.127\ J/g^{\circ} C$ .

4 0

2 years ago

According to the law of conservation of matter, the number of ________ is not changed by a chemical reaction. A. molecules B. at

laila [671]

The answer is B. atoms

7 0

2 years ago

Read 2 more answers

A 1 mol sample of gas has a temperature of 225K, a volume of 3.3L, and a pressure of 500 torr. What would the temperature be if

serg [7]

<h3>Answer:</h3>

78.75 K

<h3>Explanation:</h3>

<u>We are given;</u>

Initial pressure, P₁ = 500 torr
Initial temperature,T₁ = 225 K
Initial volume, V₁ = 3.3 L
Final volume, V₂ = 2.75 L
Final pressure, P₂ = 210 torr

We are required to calculate the new temperature, T₂

To find the new temperature, T₂ we are going to use the combined gas law;
According to the combined gas law;

P₁V₁/T₁ = P₂V₂/T₂

We can calculate the new temperature, T₂;

Rearranging the formula;

T₂ =(P₂V₂T₁) ÷ (P₁V₁)

= (210 torr × 2.75 L × 225 K) ÷ (500 torr × 3.3 L)

= 78.75 K

Therefore, the new volume of the sample is 78.75 K

5 0

3 years ago

When trying to clean muddy, dirty river water, which method would work

Llana [10]

When trying to clean muddy, dirty river water, FILTRATION would work best

6 0

2 years ago

Formaldehyde is a carcinogenic volatile organic compound with a permissible exposure level of 0.75 ppm. At this level, how many

Flauer [41]

Answer : The amount of formaldehyde permissible are, $5.4\times 10^{-6}g$

Explanation : Given,

Density of air = $1.2kg/m^3=1.2g/L$ $(1kg/m^3=1g/L)$

First we have to calculate the mass of air.

$\text{Mass of air}=\text{Density of air}\times \text{Volume of air}$

$\text{Mass of air}=1.2g/L\times 6.0L$

$\text{Mass of air}=7.2g$

Now we have to calculate the amount of formaldehyde.

Permissible exposure level of formaldehyde = 0.75 ppm = $\frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}$

Amount of formaldehyde in 7.2 g of formaldehyde = $7.2g\times \frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}$

Amount of formaldehyde in 7.2 g of formaldehyde = $5.4\times 10^{-6}g$

Thus, the amount of formaldehyde permissible are, $5.4\times 10^{-6}g$

8 0

3 years ago