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Arturiano [62]
3 years ago
11

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

metal (M). Please provide your answer in 2 decimal places.
a. How many moles of oxygen atoms are found in 4.00 g of O2?
b. How many atoms of hydrogen are found in 0.400 mol of H2S?
c. How many hydrogen atomsare present in 0.235 g of NH3?
Chemistry
1 answer:
Damm [24]3 years ago
7 0

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!

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A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c
Zarrin [17]

Answer:

C_{alloy}=0.497\frac{J}{g\°C}

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

Q_{allow}=-(Q_{water}+Q_{Al})

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})

Then, we solve for specific heat of the metallic alloy to obtain:

C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}

Thereby, we plug in the given data to obtain:

C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}

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Answer:

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Explanation:

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3 years ago
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density of a piece of ebony wood is 1.174 g/ml. what is the volume in quarts of a 2.1549 lb piece of this ebony wood?
lapo4ka [179]

0.88qt

Explanation:

Given parameters:

Density of ebony wood = 1.174g/ml

Mass of wood = 2.1549lb

Unknown:

Volume of the wood in quarts = ?

Solution:

Density is defined as the mass per unit volume of a substance.

Mass is the quantity of matter the wood contains.

Volume is the space it occupies.

To solve this problem, convert the mass form pounds into grams and plug into the density equation to find volume.

Then convert the volume from mL to quarts.

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 Volume = \frac{mass}{density}

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learn more:

Density brainly.com/question/8441651

#learnwithBrainly

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