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ollegr [7]
3 years ago
8

Which element is the least reactive

Chemistry
1 answer:
Murljashka [212]3 years ago
6 0

Answer:

air

Explanation:

air becuae aang is the avatar

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Which compound are ionic and which are covalent? (N2) (CCl4) (SiO2) (AlCl3) (CaCl2) (LiBr)
muminat
Covalent compounds: N2, CCl4, SiO2 and AlCl3.

Ionic compounds: CaCl2 and LiBr.

Hope this helps!
3 0
3 years ago
Your classmate is viewing a sample using high power and is about to refocus using the coarse adjustment knob. What would you rec
Maurinko [17]

Answer:

Using the coarse adjustment knob of the microscope in high power may lead to the breaking of the slide if adjusted and raised the slide too much which can damage the sample as well as the high power lens.

In this case, I would recommend using the fine adjustment knob and moving away from the end of the viewing area of the microscope so there would no collision take place. The fine adjustment will help to get a clear image.

7 0
2 years ago
An indicator is_____ Select all that apply.
crimeas [40]

Answer: C

Explanation:

Indicators are organic weak acids or bases with complicated structures.

5 0
2 years ago
A particular experiment requires 13.5 µl hydrochloric acid solution. if 250 ml hydrochloric acid solution is available, how many
ladessa [460]
The amount of HCl required for one experiment - 13.5 µl 
the volume in terms of L - 13.5 x 10⁻⁶ L
the volume of HCl available - 0.250 L 
since one experiment uses up - 13.5 x 10⁻⁶ L 
then number of experiments - 0.250 L / 13.5 x 10⁻⁶ L  = 1.8 x 10⁴ times 
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6 0
3 years ago
Read 2 more answers
Please help me solve this!
yulyashka [42]

Answer : The image is attached below.

Explanation :

For O_3:

Molar mass, M = 48 g/mol

Mass, m = 24 g

Moles, n = \frac{m}{M}=\frac{24g}{48g/mol}=0.5mol

Number of particles, N = n\times 6.022\times 10^{23}=0.5\times 6.022\times 10^{23}=3.0\times 10^{23}

For NH_3:

Molar mass, M = 17 g/mol

Mass, m = 170 g

Moles, n = \frac{m}{M}=\frac{170g}{17g/mol}=10mol

Number of particles, N = n\times 6.022\times 10^{23}=10\times 6.022\times 10^{23}=6.0\times 10^{24}

For F_2:

Molar mass, M = 38 g/mol

Mass, m = 38 g

Moles, n = \frac{m}{M}=\frac{38g}{38g/mol}=1mol

Number of particles, N = n\times 6.022\times 10^{23}=1\times 6.022\times 10^{23}=6.0\times 10^{23}

For CO_2:

Molar mass, M = 44 g/mol

Moles, n = 0.10 mol

Mass, m = n\times M=0.10mol\times 44g/mol=4.4g

Number of particles, N = n\times 6.022\times 10^{23}=0.10\times 6.022\times 10^{23}=6.0\times 10^{22}

For NO_2:

Molar mass, M = 46 g/mol

Moles, n = 0.20 mol

Mass, m = n\times M=0.20mol\times 46g/mol=9.2g

Number of particles, N = n\times 6.022\times 10^{23}=0.20\times 6.022\times 10^{23}=1.2\times 10^{23}

For Ne:

Molar mass, M = 20 g/mol

Number of particles = 1.5\times 10^{23}

Moles, n = \frac{N}{6.022\times 10^{23}}=\frac{1.5\times 10^{23}}{6.022\times 10^{23}}=0.25mol

Mass, m = n\times M=0.25mol\times 20g/mol=5g

For N_2O:

Molar mass, M = 44 g/mol

Number of particles = 1.2\times 10^{24}

Moles, n = \frac{N}{6.022\times 10^{23}}=\frac{1.2\times 10^{24}}{6.022\times 10^{23}}=1.9mol

Mass, m = n\times M=1.9mol\times 44g/mol=83.6g

For unknown substance:

Number of particles = 3.0\times 10^{23}

Mass, m = 8.5 g

Moles, n = \frac{N}{6.022\times 10^{23}}=\frac{3.0\times 10^{23}}{6.022\times 10^{23}}=0.50mol

Molar mass, M = \frac{m}{n}=\frac{8.5g}{0.50mol}=17g/mol

The substance is NH_3.

3 0
2 years ago
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