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REY [17]
4 years ago
9

I bet that outfit is made of Copper and Tellurium, because it is so ________.

Chemistry
2 answers:
Aleonysh [2.5K]4 years ago
7 0

THE SYMBOLS!!!

CuTe

LOLOLOL

aleksklad [387]4 years ago
4 0

Answer:

CuTe

Explanation:

The symbol for the element copper is Cu, and the symbol for tellurium is Te. The two symbols together form the word CuTe (cute).

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HELP PLEASE! :( 
BabaBlast [244]

Answer:

Part 1: W = 116 Y = 163

Part 2: Since 232 is the mailing point of 2 kg then you would divide 232 by 2 to get the melting point for 1 kg, the same with Y.

6 0
3 years ago
Read 2 more answers
1. Most of the earth's surface is covered with
frozen [14]

Answer: water

Explanation:if you look at a globe most of it is blue and blue on a globe is water so that means water covers most of the earth

5 0
3 years ago
In an aqueous solution at 25°C, if [H30+] = 3.3 * 10^4 M, then [OH-] is:
sleet_krkn [62]

Answer:

[OH-] = 3.0 x 10^-19 M

Explanation:

[H3O+][OH-] = Kw

Kw = 1.0 x 10^-14

[H3O+][OH-] = 1.0 x 10^-14

[OH-] = 1.0 x 10^-14 / 3.3 x 10^4 = 3.0 x 10^-19

3 0
3 years ago
A beaker was half filled with freshly distilled H2O and placed on a hot plate. As the temperature of the water reached 100°C, vi
jeka57 [31]

Answer:

If anything but H2O were found, it would be evidence of a chemical change. Therefore, 4 is the only viable evidence that it was a physical change of H2O liquid to H2O gas.

4 0
3 years ago
Read 2 more answers
A metal, M, forms an oxide having the formula MO2 containing 59.93% metal by mass. Determine the atomic weight in g/mole of the
Damm [24]

Answer:

See solution.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up the formula for the calculation of the by-mass percentage of the metal:

\%  M=\frac{m_M}{m_M+2*m_O}*100 \%\\\\59.93\%  =\frac{m_M}{m_M+32.00}*100 \%

Thus, we solve for the molar mass of the metal to obtain:

59.93\% (m_M+32.00) =m_M*100 \%\\\\m_M*59.93\% +1917.76\% =m_M*100 \%\\\\m_M=47.86g/mol

For the subsequent problems, we proceed as follows:

a.

4.00gO_2*\frac{1molO_2}{32.00gO_2}=0.125molO_2

b.

0.400molH_2S*\frac{2molH}{1molH_2S}*\frac{6.022x10^{23}atomsH}{1molH}=4.82x10^{23}atomsH

c.

0.235gNH_3*\frac{1molNH_3}{17.03gNH_3} *\frac{3molH}{1molNH_3}*\frac{6.022x10^{23}atomsH}{1molH}=2.49x10^{22}atomsH

Regards!

7 0
3 years ago
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