Lambda=?
c=299,792,458
f=2x10to the 12th Hz
lambda=c/f
how do I solve this?

Answers

Answer 1

Answer: $\lambda=(c)/(f)\n\nc=299729458(m)/(s)\approx3*10^(8)(m)/(s)\n\nf=2*10^(12)Hz=2*10^(12)(1)/(s)\n\n\lambda=(3*10^(8)(m)/(s))/(2*10^(12)(1)/(s))=1,5*10^(8-12)*(m)/(s)*s=1,5*10^(-4)m$

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How many moles of water (H2O) would be produced if 10 moles of O2 gas reacted completely with excess H2 gas?2 H2(g) + O2(g) → 2 H2O(g)

A. 5
B. 10
C. 20
D. 30

Answers

Answer:

C. 20

Explanation:

Use the equation to know that 1 mol of O2 is equal to 2 mol H2O.

Now you will start with the data, this is 10 mol O2 and you obtain:

10 mol O2 x 2 mol H2O = 20 mol H2O

1 mol O2

2 H2(g) + O2(g) → 2 H2O(g)

according to the reaction:
1 mole of O2-----------------2 moles of H2O
10 moles of O2--------------------- x
x = 20 mole of H2O

answer: C

For the reaction:6Li(s)+N2(g)→2Li3N(s)
Determine:
the mass of N2 needed to react with 0.536 moles of Li.
the number of moles of Li required to make 46.4 g of Li3N.
the mass in grams of Li3N produced from 3.65 g Li.
the number of moles of lithium needed to react with 7.00 grams of N2.

Answers

Explanation:

$6Li(s)+N_2(g)\rightarrow 2Li_3N(s)$

1. Mass of $N_2$ needed to react with 0.536 moles of Li.

According to reaction, 6 moles of Li reacts with 1 mol of $N_2$ .

Then 0.536 moles of Li will react with:

$(1)/(6)* 0.536$ moles of $N_2$ that is 0.0893 moles.

Mass of $N_2[tex] gas needed:</strong></p><p><strong>[tex]=28 g/mol* 0.0893 mol=2.5004 g$

2.The number of moles of Li required to make 46.4 g of $Li_3N$

Moles of $Li_3N=(46.4 g)/(35 g/mol)=1.3257 mol$

According to reaction the 2 moles of $Li_3N$ are produced from 6 moles of Li.

Then 1.3257 moles of $Li_3N$ will produced from:

$(6)/(2)* 1.3257=3.9771 moles$

3.9771 moles of lithium will needed.

3. The mass in grams of $Li_3N$ produced from 3.65 g Li.

Moles of Li $=(3.65 g)/(7 g/mol)=0.5214 moles$

According to reaction, 6 moles of Li gives 2 moles of $Li_3N$

Then 0.5214 moles of Li will give $(2)/(6)* 0.5214$ that is 0.1738 moles of $Li_3N$ .

Mass of $Li_3N=0.1738 mole* 35 g/mol=6.083 g$

6.083 grams of $Li_3N$ will be produced.

4. The number of moles of lithium needed to react with 7.00 grams of $N_2$ .

Moles of $N_2=(7.00 g)/(28 g/mol)=0.25 moles$

1 mol of $N_2$ reacts with 6 mol of Li

Then, 0.25 moles of $N_2$ will ftreact with :

$6* 0.25 moles=1.5 moles$ of lithium

1.5 moles of Li will be needed.

The Hadley , Ferrel , And polar cells are all patterns of

Answers

The Hadley, Ferrel, and Polar cells are all large scale atmospheric circulatory systems.

What Kelvin temperature is equal to -33DC? A) 306K B) -33K C) 240K D) 33K

Answers

Final Answer:

$\boxed{\boxed{\huge\textsf{$\rm -33^(\,\circ)C=240.15\,K$}}}$

Kelvin Absolute Temperature:

The kelvin, symbol K, is a unit of measurement for temperature. The Kelvin scale is an absolute scale, which is defined such that 0 K is absolute zero, the lowest temperature possible on Earth.

Historically developed from the Celsius scale, the conversion is thus:

$\boxed{\Large\textsf{$\rm 0^(\,\circ)C=273.15\,K$}}$

And absolute zero, 0 K = -273.15° C, the lowest temperature possible.

Knowing this, we can solve the question as:

$\Large\textsf{$\rm -33^(\,\circ)C=(-33+273.15)\,K$}\n\n\boxed{\boxed{\Large\textsf{$\therefore\rm -33^(\,\circ)C=240.15\,K$}}}$

$\hrulefill$

To learn more about the Kelvin Absolute Temperature Scale:

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1 ℃ = 273.15 K.

The Kelvin temperature equal to -33℃ is 240 K.

NaOH + H2SO4 =NaOH + HNO3 =
Cu(OH)2 + HCl =
H2SO4 =
HNO3 =

Answers

$2NaOH+H_2SO_4\Rightarrow\ Na_2SO_4+2H_2O\n NaOH+HNO_3\Rightarrow\ NaNO_3+H_2O\n Cu(OH)_2+2HCl\Rightarrow\ CuCl_2+2H_2O\n Cu(OH)2+H_2SO_4\Rightarrow\ CuSO_4+2H_2O\n Cu(OH)_2+2HNO_3\Rightarrow\ Cu(NO_3)_2+2H_2O$

Calculate the molar mass of cacl2

Answers

Answer:

$\boxed {\boxed {\sf 110.98 \ g/mol}}$

Explanation:

The molar mass is the mass of a substance in grams per mole.

To find it, add the mass of each element in the compound. These masses can be found on the Periodic Table.

The compound given is:

$CaCl_2$

The compound has 1 Ca (calcium) and 2 Cl (chlorine).

Mass of Calcium

The molar mass of calcium is 40.08 g/mol
There is only one atom of Calcium in CaCl₂, so the number above is what we will use.

Mass of Chlorine

The molar mass of chlorine is 35.45 g/mol
There are two atoms of chlorine in CaCl₂, therefore we need to multiply the molar mass by 2.
35.45 * 2= 70.9 g/mol

Molar Mass of CaCl₂