Through what process do humans and animals get the energy they need?

Chemistry

1 answer:

xenn [34]3 years ago

8 0

Answer:

C heat energy

Explanation:

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Explain with example that law of conservation of mass is valid for chemical reactions.

Hoochie [10]

The law of conservation of mass states that in a closed system, the mass of the system cannot change over time. Look at our example of the candle in the closed room. ... If you massed the reactants oxygen and wax, it would equal the mass of the products water and carbon dioxide.

6 0

3 years ago

If an enclosure of 0.432 L has a partial pressure of O2 of 3.4×10−6 torr at 28 ∘C, what mass of magnesium will react according t

Mama L [17]

Answer:

184620169 g.

We can say that this mass corresponds to 184.6 ton

Explanation:

This is the reaction: 2Mg (s) + O₂(g) → 2MgO (s)

We have the data of O₂, so let's find out the moles with the Ideal Gases Law.

First of all, we convert pressure to atm

760 Torr / 3.4×10⁻⁶ Torr = 223529.4 atm

P .V = n . R . T

223529.4 atm . 0.432L = n . 0.082 L.atm/mol.K . 301K

(223529.4 atm . 0.432L) / (0.082 L.atm/mol.K . 301K) = 3798769 moles

Gosh!, These are a lot of moles.

Ratio is 1:2. So with 3798769 moles we would need the double of moles of Mg. → 3798769 mol .2 = 7597538 moles

Let's convert this moles into mass ( mol . molar mass)

7597538 mol . 24.3 g/mol = 184620169 g

We can convert this mass to ton

1 ton = 1×10⁶ g

184620169 g / 1×10⁶ g = 184.6 ton

5 0

4 years ago

At a certain temperature the rate of this reaction is first order in HI with a rate constant of :0.0632s2HIg=H2g+I2g Suppose a v

olya-2409 [2.1K]

Answer:

$\boxed{\text{28.0 s}}$

Explanation:

Whenever a question asks you, "How long does it take to reach a certain concentration?" or something like that, you must use the appropriate integrated rate law expression.

The integrated rate law for a first-order reaction is

$\ln \left (\dfrac{[A]_{0}}{[A]} \right ) = kt$

Data:

[A]₀ = 1.28 mol·L⁻¹

[A] = 0.17 [A]₀

k = 0.0632 s⁻¹

Calculation:

$\begin{array}{rcl}\ln \left (\dfrac{[A]_{0}}{0.170[A]_{0}} \right ) & = & 0.0632t\\\\\ln \left (5.882) & = & 0.0632t\\1.772 & = & 0.0632t\\\\t & = & \dfrac{1.772}{0.0632}\\\\t & = & \textbf{{28.0 s}}\\\end{array}\\\text{It will take } \boxed{\textbf{28.0 s}} \text{ for [HI] to decrease to 17.0 \% of its original value.}$