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Sonja [21]
3 years ago
5

What is the percent composition by mass of hydrogen in nh4hco3

Chemistry
1 answer:
Simora [160]3 years ago
6 0

<u>Answer:</u> The percent composition by mass of hydrogen in given compound is 6.33 %

<u>Explanation:</u>

We are given:

A chemical compound having chemical formula of NH_4HCO_3

It is made up by the combination of 1 nitrogen atom, 5 hydrogen atoms, 1 carbon atom and 3 oxygen atoms

To calculate the percentage composition by mass of hydrogen in the compound, we use the equation:

\%\text{ composition of Hydrogen}=\frac{\text{Mass of hydrogen}}{\text{Mass of compound}}\times 100

Mass of compound = [(1\times 14)+(5\times 1)+(1\times 12)+(3\times 16)]=79g/mol

Mass of hydrogen = (5\times 1)=5g/mol

Putting values in above equation, we get:

\%\text{ composition of Hydrogen}=\frac{5g/mol}{79g/mol}\times 100=6.33\%

Hence, the percent composition by mass of hydrogen in given compound is 6.33 %

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6 0
3 years ago
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(a) the temperature on a warm summer day is 87 °f. what is the temperature in °c? (b) many scientific data are reported at 25
olchik [2.2K]

Answers:

              (a)  30.55 °C

              (b) 298 K and 77°F

              (c)  204.44 °C and 477.44 K

              (d)  -320.8 °F and -196 °C

Explanation:

Converting °C into °F;

                                   °F  =  °C × 1.8 + 32

Converting °F into °C;

                                   °C  =  °F - 32 ÷ 1,8

Converting °C into K;

                                   K  =  °C + 273

Converting K into °C;

                                   °C  =  K - 273

8 0
3 years ago
A 31.1 g wafer of pure gold, initially at 69.3 _c, is submerged into 64.2 g of water at 27.8 _c in an insulated container. what
KIM [24]
Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water 

Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.

From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water

At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
      = (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)- 
      = 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
       = (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
       = 268.356(T - 27.8)

Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C

Answer: 28.4 °C

3 0
3 years ago
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Because atoms are electrically neutral, the number of protons and _____ in an atom are equal.
Misha Larkins [42]

Answer: I believe that it's electrons please let me know if it was right or what it was!

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5 0
3 years ago
The enthalpy of fusion of solid n-butane is 4.66 kJ/mol. Calculate the energy required to melt 58.3 g of solid n-butane.
adelina 88 [10]

Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

Explanation :

First we have to calculate the moles of n-butane.

\text{Moles of n-butane}=\frac{\text{Mass of n-butane}}{\text{Molar mass of n-butane}}

Given:

Molar mass of n-butane = 58.12 g/mole

Mass of n-butane = 58.3 g

Now put all the given values in the above expression, we get:

\text{Moles of n-butane}=\frac{58.3g}{58.12g/mol}=1.00mol

Now we have to calculate the energy required.

Q=\frac{\Delta H}{n}

where,

Q = energy required

\Delta H = enthalpy of fusion of solid n-butane = 4.66 kJ/mol

n = moles = 1.00 mol

Now put all the given values in the above expression, we get:

Q=\frac{4.66kJ/mol}{1.00mol}=4.66kJ

Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

7 0
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