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3 years ago

PLZZZZZZ HELP ME I GIVE BRAINLIEST PLZ NO FILES!!!

Chemistry

2 answers:

Stella [2.4K]3 years ago

8 0

Answer: the Earth's equator

Explanation:

natka813 [3]3 years ago

3 0

Answer:

The light is most intense where it strikes Earth _ directly_______ to its surface.

Explanation:

Hope this helps :)

You might be interested in

What is the product when magnesium reacts with nitrogen? Mg(s) + N2(g) → Mg2N3(s) Mg3N(s) Mg3N2(s) MgN3(s)

bixtya [17]

Look at the periodic table to find the charge on atoms.
Magnesium is +2 and Nitrogen is -3. Since there are two nitrogen charge 2*-3 = -6 there needs to be 3 Mg then (3*2+ = 6+) to pair with the two nitrogen.
3 Mg(+2) + 2 N(-3) = Mg3N2

6 0

3 years ago

Read 2 more answers

2) Convert 2.65*10^ ^ 25 atoms of Chlorine to moles of CI

Irina18 [472]

Answer:

22 mol

Explanation:

Given data:

Number of atoms of Cl = 2.65×10²⁵ atom

Number of moles of Cl = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

2.65×10²⁵ atom × 1 mol / 6.022 × 10²³ atoms

0.44×10² mol

22 mol

7 0

3 years ago

If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

zlopas [31]

Answer:

$T_{eq}=28.9\°C$

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

$Q_{Cd}+Q_{W}=0$

Thus, we insert mass, specific heat and temperatures to obtain:

$m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0$

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

$T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}$

Now, we plug in to obtain:

$T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C$

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0

3 years ago

I need help PLEASEEEEEEEEEEEEEEE<br> 20 points baes

Kryger [21]

not sure but h = hydrogen and o = oxygen

7 0

3 years ago

Compute the percent ionic character of the interatomic bonds for the following compounds : a. TiO2 b. ZnTe c. CsCld. InSb e. MgC

sesenic [268]

Answer:

a. 63.2%

b. 11.7%

c. 73.3%

d. 0.995%

e. 55.5%

Explanation:

An ionic compound is a compound that is formed by ions, so one of the elements must donate electrons (which is the cation, the positive ion), and the other will receive these electrons (which is the anion, the negative ion).

The power of an element has to attract the electrons is called electronegativity, and so, as higher is the difference of electronegative of the elements, it is more probable that one of them will "still" the electrons and will form an ionic compound. The percent of this ionic character can be found by the Pauling's equation:

$%IC = (1 - e^{-0.25*(x_A - x_B)^2})$ *100%