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aleksandr82 [10.1K]
2 years ago
9

Name a metal and non metal which are liquid at normal pressure and temperature​

Chemistry
1 answer:
liraira [26]2 years ago
6 0

Answer:

metal : Mercury(Hg)

non metal : bromine (Br)

Explanation:

mercury is liquid at room temperature and pressure and the same as bromine

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Which of the following elements will form negative ions? Check<br> all that apply.
AleksandrR [38]

Answer:

Explanation:

Elements on the right side of the periodic table are very likely to form negative ions -- all of those except elements in the 8th or 18th column (depending on how your periodic table is numbered).

K and Mg are on the left side, so they will not form negative ions.

They give up 1 (for K) electron and 2 (for Mg) electrons which will leave plus charges for the ions.

On the other hand S and I are on the right side of the periodic table. They will take on electrons and hence be charged with a minus.

5 0
2 years ago
how can knowledge of percent composition help you as a consumer ? how can you promote responsible consumerism ?​
Nonamiya [84]

Answer:

it is a nice question....my mind tells me that the first is it use me as a good vibes and can use to anything the second i will do my best too absorbe it.

Explanation:

Hope this help...

8 0
3 years ago
The activity of a certain isotope dropped from 3200 ci to 800 ci in 24.0 years. what is the half-life of this isotope (in years)
Alexxx [7]
Ln(800/3200) = - kt
t = 24 years.
ln(0.25) = -k*24
(- 1.3863) = -k*24
1.3863  / 24 = k
0.05776 = k

ln(0.5) = -k*t
-0.6931 = - 0.05776 t
12 = t

I don't know if you can just look at the question and know the answer. If 24 years is a quarter life then is it obvious that the 1/2 life is 12 years? It might be, but the method I've used works for sure. 


3 0
3 years ago
Consider the reaction
spayn [35]

Answer:

1- 0.04 M/s.

2- 0.16 M/s.

Explanation:

  • For the reaction: 4PH₃ → P₄ + 6H₂.

<em>The rate of the reaction = - d[PH₃]/4dt = d[P₄]/dt = d[H₂]/6dt.</em>

where, - d[PH₃]/dt is the rate of PH₃ changing "rate of disappearance of PH₃".

d[P₄]/dt is rate of P₄ changing "rate of appearance of P₄".

d[H₂]/dt is the rate of H₂ changing "rate of formation of H₂" (d[H₂]/dt = 0.24 M/s).

<u><em>(a) At what rate is P₄ changing?</em></u>

∵ The rate of the reaction = d[P₄]/dt = d[H₂]/6dt.

∴ <em>rate of P₄ changing = </em>d[P₄]/dt = d[H₂]/6dt = (0.240 M/s)/(6.0) = 0.04 M/s.

<u><em>(b) At what rate is PH</em></u>₃<u><em> changing?</em></u>

∵ The rate of the reaction = - d[PH₃]/4dt = d[H₂]/6dt.

∴ <em>rate of PH</em>₃<em> changing = </em>- d[PH₃]/dt = 4(d[H₂]/6dt) = (4)(0.240 M/s)/(6.0) = 0.16 M/s.

7 0
3 years ago
Read 2 more answers
Water (10 kg/s) at 1 bar pressure and 50 C is pumped isothermally to 10 bar. What is the pump work? (Use the steam tables.) O -7
11Alexandr11 [23.1K]

Explanation:

For an isothermal process equation will be as follows.

                W = nRT ln\frac{P_{1}}{P_{2}}

It is given that mass is 10 kg/s or 10,000 g/s (as 1 kg = 1000 g). So, calculate number of moles of water as follows.

                    No. of moles = \frac{mass}{\text{molar mass}}

                                           = \frac{10000 g/s}{18 g/mol}

                                           = 555.55 mol/s

                                           = 556 mol/s (approx)

As T = 50^{o}C or (50 + 273.15) K = 323.15 K. Hence, putting the given values into the above formula as follows.

                  W = nRT ln[/tex]\frac{P_{1}}{P_{2}}[/tex]

                      = 556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times ln\frac{1}{10}    

                     = 556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times -2.303    

                     = -3440193.809 J/s

Negative sign shows work is done by the pump. Since, 1 J = 0.001 kJ. Therefore, converting the calculated value into kJ as follows.

                     3440193.809 J/s \times \frac{0.001 kJ}{1 J}

                          = 3440.193 kJ/s

                          = 3451 kJ/s (approx)

Thus, we can conclude that the pump work is 3451 kJ/s.

6 0
3 years ago
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