2 years ago

Need answer quick please

Chemistry

1 answer:

Nikolay [14]2 years ago

6 0

.5336
using avogadro’s law

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M. element<br>cenabrecht<br>0.5 mele of an

drek231 [11]

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Number of moles (n)=

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2 years ago

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:

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

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

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

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2 years ago

Solution made up of a

Vadim26 [7]

Answer:

solute and solvent

Explanation:

Because solutes dissolves in solvent to form a solution

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3 years ago