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kotykmax [81]
3 years ago
5

20. Which of the following are pure substances?

Chemistry
1 answer:
Juliette [100K]3 years ago
3 0
The answer is III and V
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9) What is the density of the gold rock with a mass of 386 grams and has a volume of 20 cubic centimeters?
miss Akunina [59]

Answer:

density= mass/volume

hence density of gold rock

= 386/20

=19.3 g/cc

3 0
3 years ago
In which pair of elements are both metalloids?
Aleksandr [31]

Answer:    

Sb and Te

Explanation:

there are on the stair step

3 0
3 years ago
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How many litres of fluorine gas at stp can be collected from the decomposition of 90.7 g of AuF3
7nadin3 [17]

Answer: 12 L fluorine gas at STP can be collected from the decomposition of 90.7 g of AuF_3

Explanation:

The balanced decomposition reaction is shown as

2AuF_3\rightarrow 2Au+3F_2

moles of AuF_3=\frac{\text {given mass}}{\text {Molar mass}}=\frac{90.7g}{254g/mol}=0.36moles

According to stoichiometry:

2 moles of AuF_3 gives = 3 moles of flourine gas

Thus 0.36 moles of AuF_3 gives = \frac{3}{2}\times 0.36=0.54moles of flourine gas

Using ideal gas equation :

PV=nRT

P = pressure of gas = 1 atm ( at STP)

V = Volume of gas = ?

n = moles of gas = 0.54

R = gas constant = 0.0821 L atm/Kmol

T = temperature = 273 K ( at STP)

Putting the values we get :

1atm\times V=0.54mol\times 0.0821Latm/Kmol\times 273K

V=12L

Thus 12 L fluorine gas at STP can be collected from the decomposition of 90.7 g of AuF_3

6 0
3 years ago
Describe the relationship between the mass volume and density of material
Ilia_Sergeevich [38]

Answer: For a given mass and volume, how much physical space a material takes up, of an object or substance, the density remains constant at a given temperature and pressure. The equation for this relationship is ρ = m / V in which ρ (rho) is density, m is mass and V is volume, making the density unit kg/m3.

Explanation: Hope that helps

7 0
3 years ago
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g The first step in the process is to calibrate the grating, so we know the grating spacing (the distance between the openings i
Trava [24]

Answer:

The grating spacing is  d = 1055nm

Explanation:

From the question we are told that

    The wavelength is  \lambda = 590 nm = 590 *10^{-9}m

    The angle is \theta = 34^0

    The  order of the spectrum is  n = 1

Generally in grating grating the condition for maxima   to occur is  

             d sin \theta  = n \lambda

where d is the grating spacing

     making d the subject of the formula

                  d = \frac{\lambda }{sin \theta }

        Substituting values

                  d = \frac{590 * 10^{-9}}{sin (34)}

                     d = 1055nm

8 0
3 years ago
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