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inysia [295]
3 years ago
5

Helpppppppppppp me pleaseeeeeee​

Chemistry
1 answer:
Cloud [144]3 years ago
4 0

Answer:

VX a small town called to the highest frequency and I wla kmo the

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A sample of carbon dioxide at RTP is 0.50 dm3. How many grams of carbon dioxide do we have?
prohojiy [21]

Answer:

0.924 g

Explanation:

The following data were obtained from the question:

Volume of CO2 at RTP = 0.50 dm³

Mass of CO2 =?

Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:

1 mole of gas = 24 dm³ at RTP

Thus,

1 mole of CO2 occupies 24 dm³ at RTP.

Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e

Xmol of CO2 = 0.5 /24

Xmol of CO2 = 0.021 mole

Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.

Finally, we shall determine the mass of CO2 as follow:

Mole of CO2 = 0.021 mole

Molar mass of CO2 = 12 + (2×16) = 13 + 32 = 44 g/mol

Mass of CO2 =?

Mole = mass /Molar mass

0.021 = mass of CO2 /44

Cross multiply

Mass of CO2 = 0.021 × 44

Mass of CO2 = 0.924 g.

3 0
3 years ago
3. Express the following numbers in scientific notation, making sure there is only one nonzero
cupoosta [38]

Answer:

a.

8.9 \times{10}^{ - 4}

b.

3.5470 \times {10}^{4}

Explanation:

1. Get a scientific calculator Casio fx-82ZA PLUS

2. Press shift,mode/set up, 7 and then 5

3. press whatever number you want and watch the magic happen

7 0
3 years ago
How many grams of NaOH are needed to make 500 mL of a 2.5 M NaOH solution
Veronika [31]
The number of grams  of NaOH  that are needed to make 500 ml of 2.5 M NaOH solution

calculate the number of moles =molarity x volume/1000

= 2.5 x 500/1000 = 1.25 moles

mass = moles  x molar mass of NaOH

= 1.25 x40= 50 grams of NaOH
4 0
4 years ago
Use tabulated standard electrode potentials to calculate the standard cell potential for the following reaction occurring in an
SVETLANKA909090 [29]

Answer : The standard cell potential of the reaction is, -1.46 V

Explanation :

The given balanced cell reaction is,  

3Pb^{2+}(aq)+2Cr(s)\rightarrow 2Cr^{3+}(aq)+3Pb(s)

Here, chromium (Cr) undergoes oxidation by loss of electrons and act as an anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

The standard values of cell potentials are:

Standard reduction potential of lead E^0_{[Pb^{2+}/Pb]}=-0.13V

Standard reduction potential of chromium E^0_{[Cr^{3+}/Cr]}=1.33V

Now we have to calculate the standard cell potential for the following reaction.

E^0=E^0_{cathode}-E^0_{anode}

E^0=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cr^{3+}/Cr]}

E^0=(-0.13V)-1.33V=-1.46V

Therefore, the standard cell potential of the reaction is, -1.46 V

6 0
4 years ago
How many Calories are in 6.2 * 10^4 calories? Show your work and watch the sig figs
Allushta [10]
6,200,0 is the answer to that
6 0
3 years ago
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