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2 years ago

7

A compound with the empirical formula CH2 was found to have a molar mass of approximately 112 g. Write the molecular formula of

the compound.
2Points
Show all your work. Please use correct formatting for subscripts and exponents. The math formula editor makes it easier to show work.

1 answer:

san4es73 [151]2 years ago

6 0

Answer:

C8H10

Explanation:

n (CH2) = 112

n (12 + 1 + 1) = 112

n (14) = 112

n = 8

Molecular Formula: C8H10

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A solution is formed by mixing 15.2 g KOH into

nikitadnepr [17]

Answer:

There are approximately $0.271\; \rm mol$ of formula units in that $\rm 15.2\; g$ of $\rm KOH$ (the solute of this solution.)

Explanation:

A solution includes two substances: the solute and the solvent. Note the solution here contains significantly more water than $\rm KOH$ . Hence, assume that water is the solvent (as it is in many other solutions.)

The (molar) formula mass of $\rm KOH$ is necessary for finding the number of moles of

One $\rm K$ atom,
One $\rm O$ atom, and
One $\rm H$ atom.

The formula mass of $\rm KOH$ will thus be the sum of:

The mass of one mole of $\rm K$ atoms,
The mass of one mole of $\rm O$ atoms, and
The mass of one mole of $\rm H$ atoms.

On the other hand, the mass (in grams) of one mole of atoms of an element is (numerically) the same as its relative atomic mass. The relative atomic mass data can be found on most modern periodic tables.

Relative atomic mass data from a modern periodic table:

$\rm K$ : $39.098$ .
$\rm O$ : $15.999$ .
$\rm H$ : $1.008$ .

For example, the relative atomic mass of $\rm K$ (potassium, atomic number $19$ ) is $39.098$ (3 sig. fig.) Hence, the mass of one mole of

The formula mass of $\rm KOH$ is the sum of these three masses:

$\begin{aligned}& M(\mathrm{KOH}) \\ &\approx 39.098 + 15.999 + 1.008 \\ &= 56.105\; \rm g \cdot mol^{-1}\end{aligned}$ .

The number of moles of $\rm KOH$ formula units in this $15.2\; \rm g$ sample would be:

$\begin{aligned}n &= \frac{m(\mathrm{KOH})}{M(\mathrm{KOH})} \\ &\approx \frac{15.2\; \rm g}{56.105\; \rm g \cdot mol^{-1}} \approx 0.271\; \rm mol \end{aligned}$ .

6 0

2 years ago

Consider the reaction. mc015-1.jpg How many grams of N2 are required to produce 100.0 L of NH3 at STP?

Nadusha1986 [10]

The grams of N2 that are required to produce 100.0 l of NH3 at STP

At stp 1moles = 22.4 l. what about 100.0 L of NH3

= 100 / 22.4 lx1 moles = 4.46 moles of NH3

write the reacting equation

N2+3H2 =2NH3
by use of mole ratio between N2 to NH3 which is 1:2 the moles of N2 =4.46/2 =2.23 moles of N2

mass = moles x molar mass

= 2.23moles x 28 g/mol = 62.4 grams

3 0

2 years ago

C3H8+3O2 = 3CO2+4H2O what is the enthalpy combustion

dezoksy [38]

Answer:

$\Delta H_{comb}=2043.85kJ/mol$

Explanation:

Hello there!

In this case, according to the given chemical reaction, it possible for us to set up the expression for the calculation of the enthalpy change as shown below:

$\Delta H_r=-\Delta H_{comb}=3\Delta _fH_{CO_2}+4\Delta _fH_{H_2O}-\Delta _fH_{C_3H_8}-3\Delta _fH_{O_2}$

Thus, given the values of the enthalpies of formation on the attached file, we obtain: $-\Delta H_{comb}=3(-393.5kJ/mol)+4(-241.8kJ/mol)-(-103.85kJ/mol)-3(0kJ/mol)\\\\-\Delta H_{comb}=-2043.85kJ/mol\\\\\Delta H_{comb}=2043.85kJ/mol$

Best regards!

8 0

2 years ago

A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$

3. Plot the temperature readings as a function of time.

The graphs are shown below.

6 0

2 years ago

These two questions please

zhenek [66]

Answer:

1.) 3

2.) 60 CM

Explanation:

1. Density= $\frac{MASS}{VOLUME}$ = $\frac{75}{25}$

2. Length*Width*Height=3*10*2

6 0

2 years ago

Read 2 more answers