1) the equation of the energy is: E=c*m*(t₂-t₁), where E - energy (523 J), c - unknown specific heat of copper, m - mass of this copper [kg], t₂ - the final temperature, t₁ - initial temerature;

2) the specific heat of copper is:

$c=\frac{E}{m*(t_2-t_1)}; \ => \ c=\frac{523}{0.085*(45-30)}=\frac{523}{1.275}=410.196[\frac{J}{kg*C}].$

6 0

2 years ago

During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.

Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.

it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of CH₄ needs → 2 mol of O₂

??? mol of CH₄ needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over * molar mass

= 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

3 0

2 years ago

Select all the correct images.<br> Select the atomic models that belong to the same element.

Alex_Xolod [135]

Answer:The 2nd and 3rd one.

Explanation:

It has the same number of protons but different amount of nuetrons.

7 0

2 years ago