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3 years ago

Which law is based on the graph that is shown below?

Chemistry

1 answer:

Cerrena [4.2K]3 years ago

8 0

Answer:

Boyle's Law

Explanation:

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For the problem: Mg (s) + O₂ (g) ➞ MgO (g) if 4.3 g of magnesium reacts with 83 g of oxygen (O₂), which is the limiting reagent?

Anit [1.1K]

Explanation:

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3 years ago

There is nothing that you can do to conserve energy. True or False

Firdavs [7]

I am going to say it is false.

6 0

3 years ago

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What particle decay is this? 210 83 Bi→210 84 Po

sesenic [268]

Answer:

Beta emission

Explanation:

In beta emission, a neutron is converted into a proton thereby emitting an electron and a neutrino. A neutrino is a particle that serves to balance the spins.

When a nucleus undergoes beta emission, the mass number of the parent and daughter nuclei remain the same while the atomic number of the daughter nucleus is greater than that of its parent by one unit.

Hence, in beta emission, the daughter nucleus is found one pace to the right of the parent in the periodic table.

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3 years ago

The substances that are produced after the chemical reaction are called the ___

sertanlavr [38]

Answer:

Products

Explanation:

During a chemical reaction, there are the reactants (left side), and the products (right side).

6 0

2 years ago

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The rate of disappearance of HBr in the gas phase reaction2HBr(g) → H2(g) + Br2(g)is 0.301 M s-1 at 150°C. The rate of appearanc

jok3333 [9.3K]

Answer: 0.151

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$2HBr(g)\rightarrow H_2(g)+Br_2(g$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

$Rate=-\frac{1d[HBr]}{2dt}=+\frac{1d[H_2]}{2dt}=+\frac{1d[Br_2]}{dt}$

Given: $-\frac{d[HBr]}{dt}]=0.301$

Putting in the values we get:

$\frac{0.301}{2}=+\frac{1d[Br_2]}{dt}$

$+\frac{1d[Br_2]}{dt}=0.151$

Thus the rate of appearance of $Br_2$ is 0.151

8 0

3 years ago