Cu(NO3)2 + Na2CO3 → CuCO3 + 2NaNO3

What mass of CO was used up in the reaction with an excess of oxygen gas if 24.7g of carbon dioxide is formed? 2 CO + O2 > 2

Zolol [24]

Balance Chemical Equation,
2 CO + O₂ → 2CO₂
Acc. to this reaction,
88 g (2 mole) of CO₂ was produced when = 56 g (2 mole)of CO was reacted
So,
24.7 g of CO₂ will be produced by reacting = X g of CO

Solving for X,
X = (56 g × 24.7 g) ÷ 88 g

X = 2.26 g ÷ 88 g

X = 0.0257 g of CO

Result:
0.0257 g of CO is required to be reacted with excess of O₂ to produce 24.7 g of CO₂.

7 0

3 years ago

What is 5.0553 x 10^-4 in standard notation?

Talja [164]

Answer:

0.00050553

Explanation:

when the power of ten is negative, move the decimal to the left

hope this helped!

6 0

3 years ago

Calculate the frequency of the n=2 line in the lyman series of hydrogen

Alona [7]

Answer:

Approximately $2.47\times 10^{15}\; \rm Hz$ .

Explanation:

The Lyman Series of a hydrogen atom are due to electron transitions from energy levels $n \ge 2$ to the ground state where $n = 1$ . In this case, the electron responsible for the line started at $n = 2$ and transitioned to

A hydrogen atom contains only one electron. As a result, Bohr Model provides a good estimate of that electron's energy at different levels.

In Bohr's Model, the equation for an electron at energy level $n$ (

$\displaystyle - \frac{k\, Z^2}{n^2}$ (note the negative sign in front of the fraction,)

where

$k = 2.179 \times 10^{-18}\; \rm J$ is a constant.

$Z$ is the atomic number of that atom. $Z = 1$ for hydrogen.
$n$ is the energy level of that electron.

The electron that produced the $n = 2$ line was initially at the

$\begin{aligned} &E_{n = 2} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{2^2} \cr & \approx -5.4475\times 10^{-19}\; \rm J\end{aligned}$ .

The electron would then transit to energy level $n = 1$ . Its energy would become:

$\begin{aligned} &E_{n = 1} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{1^2} \cr & \approx -2.179 \times 10^{-18} \; \rm J\end{aligned}$ .

The energy change would be equal to

$\begin{aligned}&\text{Initial Energy} - \text{Final Energy} \cr &= E_{n = 2} - E_{n = 1} \cr &= -5.4475 \times 10^{-19} - \left(-2.179 \times 10^{-18}\right) \cr & \approx 1.63425\times 10^{-18}\; \rm J \end{aligned}$ .

That would be the energy of a photon in that $n = 2$ spectrum line. Planck constant $h$ relates the frequency of a photon to its energy:

$E = h \cdot f$ , where

$E$ is the energy of the photon.
$h \approx 6.62607015\times 10^{-34}\; \rm J \cdot s$ is the Planck constant.
$f$ is the frequency of that photon.

In this case, $E \approx 1.63425 \times 10^{-18}\; \rm J$ . Hence,

$\begin{aligned} f &= \frac{E}{h} \cr &\approx \frac{1.63425\times 10^{-18}}{6.62607015\times 10^{-34}} \cr & \approx 2.47 \times 10^{15}\; \rm s^{-1}\end{aligned}$ .

Note that $1 \; \rm Hz = 1 \; \rm s^{-1}$ .

6 0

3 years ago

Write an equation that shows the reaction between acetic acid and triethylamine (CH3CH2)3N. Draw all non-bonding lone electron p

Anni [7]

Answer:

-) Acid-base reaction

-) Carboxylic acid, alcohol, alkene and ketone

Explanation:

For the reaction between acetic acid and triethylamine, we will have an acid-base reaction. Therefore a salt would be produced in this case an "ammonium quaternary salt". Also, we have to remember that on this reaction the acid is the acetic acid and the base is the triethylamine. See figure 1

For the second question, we have to check the structure of Prostaglandin E1 in which we have the functional groups:

1) Carboxylic acid

2) Alcohol

3) Alkene

4) Ketone

See figure 2.

I hope it helps!

8 0

3 years ago

Find the grams of iron in 79.2 g of Fe2O3.

Setler79 [48]

First you find the RFM (Relative Formula Mass) of Fe2O3
Fe2 = 56 x 2
= 112
O3 = 16 x 3
= 48

112 + 48 = 160

Moles = mass/RFM
   = 79.2/160
   =0.495

Then you rearrange the formula
Mass= moles x RFM
      = 0.495 x 112
      = 55.44 g
(To find mass of Iron have to use RFM of total iron in the substance example Fe2 ^^)

3 0

3 years ago