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3 years ago

How can you increase the momentum of an object?

Chemistry

1 answer:

Elena-2011 [213]3 years ago

5 0

I think its B because if u increase the mass itll have more force which will increase the momentum

You might be interested in

Elements - made up of<br> Ex:

astra-53 [7]

All elements are made up of atoms. ➢ Atoms are made up of protons, neutrons, and electrons. Two different kinds of atoms can combine to form a compound. A molecule is a combination of atoms that cannot be broken apart while still retaining the same properties as the larger substance that it is a part of.

7 0

2 years ago

Using the balanced equation for the combustion of ethane: 2C2H6 + 7O2 → 4CO2 + 6H2O, how many moles of O2 needed to produce 12 m

pickupchik [31]

Answer:

14 moles of oxygen needed to produce 12 moles of H2O.

Explanation:

We are given that balance eqaution

$2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O$

We have to find number of moles of O2 needed to produce 12 moles of H2O.

From given equation

We can see that

6 moles of H2O produced by Oxygen =7 moles

1 mole of H2O produced by Oxygen= $\frac{7}{6}$ moles

12 moles of H2O produced by Oxygen= $\frac{7}{6}\times 12$ moles

12 moles of H2O produced by Oxygen= $7\times 2$ moles

12 moles of H2O produced by Oxygen=14 moles

Hence, 14 moles of oxygen needed to produce 12 moles of H2O.

3 0

2 years ago

an engineer wishes to design a container that will hold 12.0 mol of ethane at a pressure no greater than 5.00x10*2 kPa and a tem

OleMash [197]

Answer:

The minimum volume of the container is 0.0649 cubic meters, which is the same as 64.9 liters.

Explanation:

Assume that ethane behaves as an ideal gas under these conditions.

By the ideal gas law,

$P\cdot V = n\cdot R\cdot T$ ,

$\displaystyle V = \frac{n\cdot R\cdot T}{P}$ .

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles of particles in this gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas (in degrees Kelvins.)

The numerical value of $R$ will be $8.314$ if $P$ , $V$ , and $T$ are in SI units. Convert these values to SI units:

$P =\rm 5.00\times 10^{2}\;kPa = 5.00\times 10^{2}\times 10^{3}\; Pa = 5.00\times 10^{5}\; Pa$ ;
$V$ shall be in cubic meters, $\rm m^{3}$ ;
$T = \rm 52.0 \textdegree C = (52.0 + 273.15)\; K = 325.15\; K$ .

Apply the ideal gas law:

$\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P}\\ &= \frac{12.0\times 8.314\times 325.15}{5.00\times 10^{5}}\\ &= \rm 0.0649\; m^{3} \\ &= \rm (0.0649\times 10^{3})\; L \\ &=\rm 64.9\; L\end{aligned}$ .

4 0

3 years ago

During an investigation, a student burns magnesium to form magnesium oxide. The starting mass of magnesium is measured as 21.3 g

enot [183]

Answer:

Percentage yield = 85.2%

Explanation:

Given data:

Mass of Mg = 21.3 g

Actual yield of MgO = 30.2 g

Percentage yield = ?

Solution:

Chemical equation:

2Mg + O₂ → 2MgO

Number of moles of Mg = mass/molar mass

Number of moles of Mg = 21.3 g / 24.3 g/mol

Number of moles of Mg = 0.88 mol

Now we will compare the moles of MgO with Mg.

Mg : MgO

2 : 2

0.88 : 0.88

Mass of MgO:

Mass of MgO= moles × molar mass

Mass of MgO= 0.88 mol × 40.3g/mol

Mass of MgO = 35.46 g

Actual yield of MgO = 30.2 g

Percentage yield:

Percentage yield = Actual yield/theoretical yield × 100

Percentage yield = 30.2 g/ 35.46 g × 100

Percentage yield = 85.2%

3 0

3 years ago

Guys, please, please help!!

shutvik [7]

Ookay

1) T
2) T
3) F

Hope i helped :)

6 0

3 years ago

Read 2 more answers