All categories

3 years ago

1. Energy changes occur as *

Chemistry

1 answer:

alexandr1967 [171]3 years ago

3 0

Number 3 i think is d.heat moves from an object of higher temperature to an object of lower temperature

You might be interested in

What is true of a scientific law?

Minchanka [31]

Answer: d

Explanation:

5 0

3 years ago

If 4000 g of Fe2O3 reacts, how many moles of CO are needed? (add work)

7nadin3 [17]

Answer:

75.15 mol.

Explanation:

Firstly, we need to write the balanced equation of the reaction:

Fe₂O₃ + 3CO → 2Fe + 3CO₂.

It is clear that 1.0 mole of Fe₂O₃ reacts with 3.0 moles of CO to produce 2.0 moles of Fe and 3.0 moles of CO₂.

∴ Fe₂O₃ reacts with CO with (1: 3) molar ratio.

we need to calculate the no. of moles of (4000 g) of Fe₂O₃:

no. of moles of Fe₂O₃ = mass/molar mass = (4000 g)/(159.69 g/mol) = 25.05 mol.

Using cross multiplication:

1.0 mole of Fe₂O₃ needs → 3.0 moles of CO,

∴ 25.05 mole of Fe₂O₃ needs → ??? moles of CO.

∴ The no. of moles of CO needed = (3.0 mol)(25.05 mol)/(1.0 mol) = 75.15 mol.

5 0

4 years ago

A reaction mixture initially contains 0.140 MCO and 0.140 MH2O. What will be the equilibrium concentration of CO?

Allushta [10]

Answer:

0.013 M

Explanation:

From the question, we can make the following deductions; we are given mixture that contains two compounds, that is A and B, 0.140 M CO and 0.140 M H2O respectively. Then, we are asked to find the equilibrium concentration of Carbonmonoxide,CO.

So, the equation for the reaction is given below;

CO + H2O <-----------------> CO2 + H2.

Initially: we have 0.14M of CO, 0.14M of H2O and zero (0) concentration of CO2 and H2.

At time,t = CO =0.14 - x , H2O = 0.14 - x, CO2 and H2 = x.

The above reaction consist of the forward reaction and the backward reaction.

Therefore, the equilibrium Concentration of CO;

(Since we are giving that Kc = 102). Then, Kc= [CO2][H2] ÷ [CO][H2O]. Where Kc is the equilibrium constant.

Therefore, 102 = [x^2] / [0.14 - x]^2.

==> 10.1= x/0.14 - x.

====> 0.141 - 10.1 x = x.

x + 10.1 x = 0.141.

===> 11.1 x = 0.141.

===> x = 0.141 ÷ 11.1.

===> x = 0.127 M .

Then, at time,t CO = 0.14 - x.

= 0.14 - 0.127 = 0.013 M

8 0

4 years ago

Read 2 more answers

The two isotopes of chlorine are LaTeX: \begin{matrix}35\\17\end{matrix}Cl35 17 C l and LaTeX: \begin{matrix}37\\17\end{matrix}C

Kay [80]

Answer: The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

For $_{17}^{35}\textrm{Cl}$ isotope:

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 35 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

For $_{17}^{37}\textrm{Cl}$ isotope:

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 37 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.45 amu

Putting values in equation 1, we get:

$35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.775\times 100=77.5\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.775)=0.225\times 100=22.5\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

6 0

3 years ago

An astronaut has a mass of 60kg. Calculate their weight When they are on the surface of the moon.

IceJOKER [234]

Answer:

I think 0kg because they are flying.

7 0

3 years ago