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3 years ago

14

What are some signs that a chemical change takes place when a firework explodes? Matter is destroyed Light is produced Heat is p

roduced The chemical composition is not changed

1 answer:

Radda [10]3 years ago

4 0

Well, the answer relies on all four factors, I mean on all options. So, the answer is A, B, C, and D.

You might be interested in

For a grating, how many lines per millimeter would be required for the first order diffraction line for λ = 400 nm to be observe

MA_775_DIABLO [31]

Answer:

For a grating, how many lines per millimeter would be required for the first order diffraction line for λ = 400 nm to be observed at a reflection angle of $5^{o}$ when the angle of incidence is $45^{o}$ ?

<em>The number of lines per mm is 1985 lines </em>

Explanation:

A diffraction grating is used to separate a light source with multiple wavelengths into single wavelengths of different colors using the principle of diffraction.

Calculating for the spacing between the reflected surfaces

The spacing between the reflecting surfaces can be obtained using the relationship with the wavelength as shown below;

nA = d (sin i + sin r) ....................................1

where n is the order of diffraction = 1;

A is the wavelength = 400 nm

i is the angle of incidence = $45^{o}$

r is the angle of refraction = $5^{o}$

making d the subject formula from equation 1 we have;

$d = \frac{nA}{sin i + sin r}$ ...........................2

Now we substitute the given parameters into equation 2;

$d = \frac{1(400 nm)}{(sin 45^{o} ) + (sin 5^{o} )}$

d = $\frac{400 nm}{0.707 +0.087}$

d = 503.7 nm

The spacing between reflecting surfaces is 503.7 nm

Calculating for the number of lines per millimeter

To calculate the number of spacing per millimeter we convert nanometer to millimeters, 1 mm = $10^{6}$ and 1 line = 503.7 nm;

$1 mm *\frac{1 line}{503.7 nm} *\frac{10^{6} }{mm}$

= 1985 lines

Therefore the number of lines per millimeter would be 1985 lines

6 0

3 years ago

Lead (II) nitrate and magnesium iodide solutions are mixed

sashaice [31]

The molecular equation :

MgI₂(aq) + Pb(NO₃)₂(aq) → Mg(NO₃)₂(aq) + PbI₂(s)

The net ionic equations

Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s)

<h3>Further explanation</h3>

Given

Word equation

Required

The molecular equations and the net ionic equations

Solution

The chemical equation can be expressed in terms of:

word equation
skeleton equation
balanced equation

The equation of a chemical reaction can be expressed in the equation of the ions

When a spectator ions are removed, the ionic equation is called the net ionic equation

A molecular equation is a chemical equation that has been balanced and is written in molecular (not ionic) form

The molecular equation :

MgI₂(aq) + Pb(NO₃)₂(aq) → Mg(NO₃)₂(aq) + PbI₂(s)

the ionic equation :

Mg²⁺ +2I⁻ + Pb²⁺ + 2NO₃⁻ → Mg²⁺ + 2NO₃⁻ + PbI₂

the net ionic equations

Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s)

5 0

2 years ago

Need help answering the question at the bottom of the text

Rasek [7]

Answer:

You Need to put in the picture

Explanation:

I cant see it

7 0

2 years ago

A steel container filled with H₂ gas is at a pressure of 6.5 atm and a temperature of 22°C. If the container is placed near a fu

Yuki888 [10]

Answer: The new pressure is 7.1 atm

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=6.5atm\\T_1=22^0C=(22+273)K=295K\\P_2=?\\T_2=50^0C=(50+273)K=323K$

Putting values in above equation, we get:

$\frac{6.5}{295}=\frac{P_2}{323}\\\\P_2=7.1$

Hence, the new pressure is 7.1 atm

8 0

2 years ago

Polonium- 210 , Po210 , decays to lead- 206 , Pb206 , by alpha emission according to the equation Po84210⟶Pb82206+He24 If the ha

elixir [45]

Answer:

0.269 g

Explanation:

$Po_{210} ^{84}$ ⟶ $Pb_{206} ^{82}$ + $+ He_{4} ^{2}$

Data:

Half-life of $Po_{210} ^{84}$ (T(1/2)) = 138.4 days

Mass of PoCl4 = 561 mg (0,561 g) and molecular weight of PoCl4 = 350. 79 g/mol

Time = 338.8 days

Isotopic masses

$Po_{210} ^{84}$ = 209.98 g/mol

$Pb_{206} ^{82}$ = 205.97 g/mol

Concepts

Avogadro’s number: This is the number of constituent particles that are contained in a mol of any substance. These constituted particles can be atoms, molecules or ions). Its value is 6.023*10^23.

The radioactive decay law is

N=Noe^(-λt)

Where:

No = number of atoms in t=0

N = number of atoms in t=t (now) in this case t=338.8 days

λ= radioactive decay constant

The radioactive constant is related to the half-life by the next equation

λ= $\frac{ln 2}{t(1/2)}$

so

λ= $\frac{ln2}{138.4 days}$ =0,005008 days^(-1)

No (Atoms of $Po_{210} ^{84}$ in t=0)

To get No we need to calculate the number of atoms of $Po_{210} ^{84}$ in the initial sample. We have a sample of 0,561 g of PoCl4. If we get the number of moles of PoCl4 in the sample, this will be the number of moles of $Po_{210} ^{84}$ in the initial sample.

This is:

$\frac{0,561 g of PoCl4}{350. 79 g of PoCl4 /mol}$ = 0,001599 mol of PoCl4

This is the number of mol of $Po_{210} ^{84}$ in the initial sample.

To get the number of atoms in the initial sample we use the Avogadro’s number = 6.023*10^23

0,001599 mol of $Po_{210} ^{84}$ * 6.023*10^23 atoms/ mol of $Po_{210} ^{84}$ = 9.632 *10^20 atoms of $Po_{210} ^{84}$

Atoms after 338.8 days

We use the radioactive decay law to get this value

N=Noe^(-λt)

N=9.632*10^20 e^(-0,005008 days^(-1) * 338.8 days) =1.765*10^20

This is the number of atoms of $Po_{210} ^{84}$ in the sample after 338.8 days has passed

The number of atoms $Po_{210} ^{84}$ transformed is equal to the number of atoms of $Pb_{206} ^{82}$ produced.

The number of atoms of $Po_{210} ^{84}$ transformed is No - N

9.632 *10^20 – 1.765 *10^20 = 7.866*10^20

So, 7.866*10^20 is the number of atoms of $Pb_{206} ^{82}$ produced

We can get the mass with the Avogadro’s number

(7.866*10^20 atoms of $Pb_{206} ^{82}$ ) / ( 6.023*10^23 atoms of $Pb_{206} ^{82}$ / mol of $Pb_{206} ^{82}$ = 0.001306 moles of $Pb_{206} ^{82}$

This number of moles have a mass of:

(0,001306 moles of $Pb_{206} ^{82}$ )* (205.97 g of $Pb_{206} ^{82}$ /mol of $Pb_{206} ^{82}$ ) = 0.269 g

3 0

3 years ago