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3 years ago

Which of the following are examples of plasmas?

Chemistry

1 answer:

Burka [1]3 years ago

4 0

Answer:

i will tell some examples of plasmas they are:

1.lightning

2.solar wind

3.welding arcs

4.stars(including the sun)

5.the earths's ionosphere

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Are humans pro-karyotic or eukaryotic?

barxatty [35]

Answer:

eukaryotic

Explanation:

all human cells—including those found in the brain, the heart, the muscles, and so on—are also eukaryotic.

3 0

3 years ago

How many Joules are released to cool 250.0 grams of liquid water from 100°C to 0°C? The specific heat of water is 4.180 J/g.C.

nlexa [21]

$\bold{\huge{\orange{\underline{ Solution}}}}$

$\bold{\underline{ Given :- }}$

We have 250g of liquid water and it needs to be cool at temperature from 100° C to 0° C
Specific heat of water is 4.180J/g°C

$\bold{\underline{ To \: Find :- }}$

We have to find the total number of joules released.

$\bold{\underline{ Let's \:Begin:- }}$

We know that,

Amount of heat energy = mass * specific heat * change in temperature

That is, 

$\sf{\red{ Q = mcΔT }}$

Subsitute the required values in the above formula :-

$\sf{ Q = 250 × 4.180 ×(0 - 100 )}$

$\sf{ Q = 250 × 4.180 × - 100 }$

$\sf{ Q = 250 × - 418}$

$\sf{\pink{ Q = - 104,500 J }}$

Hence, 104,500 J of heat is released to cool 250 grams of liquid water from 100° C to 0° C.

$\bold{\underline{ Now :- }}$

We have to tell whether the above process is endothermic or exothermic :-

Here, In the above process ΔT is negative and as a result of it Q is also negative that means above process is Exothermic

Exothermic process :- It is the process in which heat is evolved .
Endothermic process :- It is the process in which heat is absorbed .

4 0

2 years ago

How does the size of a nanoparticle compare with the size of an atom?

RSB [31]

A nanoparticle is larger than an atom. A nanoparticle is usually made from a few hundred atoms. These particles range from 1 nanometers to 100 nanometers. On the other hand an atom ranges from 0.1 nanometers to 105 nanometers. Using the sizes above, one can clearly see and understand that an atom is smaller.

6 0

3 years ago

Which of the following represents the ground state electron configuration for the Mn3+ ion? (Atomic number Mn = 25) 1s2 2s22p6 3

V125BC [204]

Answer:

$1s^22s^22p^63s^23p^63d^{4}$

Explanation:

Manganese is the element of group 7 and forth period. The atomic number of Manganese is 25 and the symbol of the element is Mn.

The electronic configuration of the element, manganese is -

$1s^22s^22p^63s^23p^63d^{5}4s^2$

To form $Mn^{3+}$ , it will lose 3 electrons from the valence electrons and thus the configuration of the ion is:-

$1s^22s^22p^63s^23p^63d^{4}$

8 0

3 years ago

Read 2 more answers

Suppose 6.63g of zinc bromide is dissolved in 100.mL of a 0.60 M aqueous solution of potassium carbonate. Calculate the final mo

Alenkasestr [34]

Answer:

[Zn²⁺] = 4.78x10⁻¹⁰M

Explanation:

Based on the reaction:

ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)

The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:

1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]

We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:

Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:

6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺

Moles K₂CO₃ = Moles CO₃²⁻:

0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻

Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =

0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]

Replacing in Ksp expression:

1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]

4 0

3 years ago