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3 years ago

Do all living things give off co2?

Chemistry

1 answer:

Yuki888 [10]3 years ago

8 0

Yes. This is mainly due to chemical processes occurring in the body of an organism/living thing.

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Given the chemical reaction Fe3O4 + H2 ⟶ Fe + H2O, identify the coefficient of H2O in the balanced equation.

Kay [80]

Add 4H2 in the reactant side, that will give you 4H2O in the product side.

5 0

3 years ago

Read 2 more answers

How is Hydrogen in heavy water different from hydrogen in normal water

soldier1979 [14.2K]

Answer: An oxygen atom in heavy water has an extra neutron. A hydrogen atom in heavy water has an extra proton.

Explanation:

7 0

3 years ago

What is the specific heat of a substance if 25 grams rises in temperature from 10 degrees Celsius to 25 degrees with the additio

Marina86 [1]

Answer:

0.75 cal/g°c

Explanation:

for specific heat we have formula:

Amount of heat absorbed or released = mass x specific heat of a substance x change in temperature.

ΔQ=m x c x ΔT

where c= specific heat

m= mass of a substance

ΔT = total temperature

ΔQ = Amount of heat

so for specific heat,

c= ΔQ/mxΔT

c= 280/25x (25-10)

c= 280/375

c= 0.75 cal/g°c

4 0

4 years ago

There are approximately 15 milliliters (mL) in 1 tablespoon (tbsp). Which expression can be used to find the approximate number

Helga [31]

You know 1 tbsp is 15 milliliters so you would multiply 15 and 3 together. 15 x 3 =45
So there are 45 milliliters in 3 tbsp.

8 0

3 years ago

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Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the abo

Romashka-Z-Leto [24]

Answer:

$\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}$

Explanation:

Volume of a cone:

$\displaystyle V=\frac{1}{3} \pi r^2 h$

We have $\displaystyle \frac{dV}{dt} = \frac{10 \ cm^3}{sec}$ and we want to find $\displaystyle \frac{dh}{dt} \Biggr | _{h\ =\ 6}= \ ?$ when the height is 2 cm.

We can see in our equation for the volume of a cone that we have three variables: V, r, and h.

Since we only have dV/dt and dh/dt, we can rewrite the equation in terms of h only.

We are given that the height of the cone is 1/5 the radius at any given time, 1/5r, so we can write this as r = 5h.

Plug this value for r into the volume formula:

$\displaystyle V =\frac{1}{3} \pi (5h)^2 h$
$\displaystyle V =\frac{1}{3} \pi \ 25h^3$

Differentiate this equation with respect to time t.

$\displaystyle \frac{dV}{dt} =\frac{25}{3} \pi \ 3h^2 \ \frac{dh}{dt}$
$\displaystyle \frac{dV}{dt} =25 \pi h^2 \ \frac{dh}{dt}$

Plug known values into the equation and solve for dh/dt.

$\displaystyle 10 = 25 \pi (2)^2 \ \frac{dh}{dt}$
$\displaystyle 10 = 100 \pi \ \frac{dh}{dt}$

Divide both sides by 100π to solve for dh/dt.

$\displaystyle \frac{10}{100 \pi} = \frac{dh}{dt}$
$\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}$

The height of the cone is increasing at a rate of 1/10π cm per second.

7 0

3 years ago