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3 years ago

What is the mass of hydrogen in 50.0 g of propane, C3H8?

Chemistry

2 answers:

Oksanka [162]3 years ago

6 0

The atomic molar mass of C is 12 and of H is 1. Hence the mass ratio of H to the molecule is 8/(8+36) = 2/11

Since 50*2/11 ≈ 9, the answer is<u><em> B.</em></u>

Phoenix [80]3 years ago

6 0

Answer: 9.1 g

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For $C_3H_8$

Given mass = 50.0 g

Molar volume of $C_3H_8$ = 44 g/mol

Putting values in above equation, we get:

$\text{Moles of}C_3H_8=\frac{50}{44}=1.14mol$

According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of $C_3H_8$ contains 8 moles of hydrogen

1.14 moles of $C_3H_8$ contains = $\frac{8}{1}\times 1.14=9.1$ moles of hydrogen

Mass of hydrogen= $moles\times {\text {molar mass}}=9.1\times 1.0=9.1 g$

Thus the mass of hydrogen is 9.1 grams.

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3 years ago

Write 0.00000009345 in Engineering Notation with 3 significant figures

melisa1 [442]

Answer:

$93.43\times 10^{-9}$

Explanation:

Scientific notation is the way of writing numbers which are either large or small. The number is written in the scientific notation when the number is between 1 and 10 and then multiplied by the power of 10. Engineering notation is the same version of the scientific notation but the number can be between 1 and 1000 and in this exponent of the ten is divisible by three.

For example, $1000^2$ is to be written as $10^6$ in engineering notation.

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0.00000009345 can be written as $93.425\times 10^{-9}$

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3 years ago

Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm

8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

$\frac{dV}{dt}=10\frac{cm^3}{s}$

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

$\frac{dh}{dt} = ?*\frac{dV}{dt}$

Of course, $?=\frac{dh}{dV}$ .

Now, since the volume of a cone is $V=\pi r^2h/3$ and the ratio $r/h=15/20=3/4$ or $r=3/4h$ , the volume becomes:

$V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3$

We proceed to its differentiation:

$\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}$

Then, we compute $\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}$

Finally, at h=2:

$\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}$

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4 years ago

Please help!!!!

Nat2105 [25]

Answer:

When an atom gains or loses energy, the energy of an electron can change.

An electron in an atom can move from one energy level to another when the atom gains or loses energy.

Explanation:

The possible energies that electrons in an atom

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https://1.cdn.edl.io/tTlW7xRtvD62xSe7RcZlJr7kSR7XsL93akcgJkbGJBNNcpwY.pdf

*this link can also help* brainly ist plz

8 0

3 years ago

Read 2 more answers

What is the empirical formula for a compound that contains 10.89% magnesium 31.77% chloride and 57.34% oxygen

Blizzard [7]

Answer: $Mg_{1}Cl_{2}O_{8}$

Explanation:

If percentage are given then we are taking total mass is 100 grams.So, the mass of each element is equal to the percentage given.

Mass of Mg = 10.89 g

Mass of Cl = 31.77 g

Mass of O = 57.34 g

Step 1 : convert given masses into moles.

Moles of Mg= $\frac {\text{ given mass of Mg}}{\text{ molar mass of Mg}}= \frac {10.89g}{24g/mole}=0.45moles$

Moles of Cl = $\frac {\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac {31.77g}{35.5g/mole}=0.89moles$

Moles of O = $\frac {\text{ given mass of O}}{\text{ molar mass of O}}=\frac {57.34g}{16g/mole}=3.58moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Mg = $\frac {0.45}{0.45}=1$

For Cl = $\frac {0.89}{0.45}=2$

For O= $\frac {3.58}{0.45}=8$

The ratio of Mg :Cl : O= 1 : 2 : 8

Hence the empirical formula is $Mg_{1}Cl_{2}O_{8}$

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3 years ago