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Xelga [282]
3 years ago
11

How can a jet create a sonic boom?

Chemistry
1 answer:
Alina [70]3 years ago
6 0
It creates a Sonic boom by travelling faster than the speed of sound.
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How do you Calculate the molarity of 250 ml of solution in which 2.7 g of MgCl2 are dissolved
IRINA_888 [86]

Answer:

Molar mass of MgCl2 is 95 g/mol

Mg = 24 g/mol and Cl = 35.5 ×2 = 71 g/mol

moles = mass given/ molar mass

= 2.7/95 = 0.028 mol

volume = 250/1000 = 0.25 dm3 (ml is the same as dm3)

molarity of MgCl2 = moles/volume

= 0.028/0.25

= 0.112 mol/dm3

3 0
2 years ago
How many total moles of ions are released when each of the following dissolves in water?
EleoNora [17]

we have a total of three times the original number (6.923 * 10**-7) moles of all ions, or 2.077 * 10**-6 moles of ions

<h3>What is aragonite-strontianite solid solution dissolution in nonstoichiometric Sr (HCO3)2 solutions?</h3>

Synthetic strontianite-aragonite solid-solution minerals were dissolved in non-stoichiometric CO2-saturated Sr(HCO3)2 and Ca(HCO3)2 solutions at 25°C.  The reactions in Sr(HCO3)2 solutions frequently become incongruent, precipitating a Sr-rich phase before attaining stoichiometric saturation. Mechanical mixes of solids approach stoichiometric saturation in terms of the least stable solid in the combination.

This surficial phase has a thickness of 0-10 atomic layers in Sr(HCO3)2 solutions and a thickness of 0-4 layers in Ca(HCO3)2 solutions and dissolves and/or recrystallizes within 6 minutes of reaction.

learn more about Sr (HCO3)2 refer

brainly.com/question/24667072

#SPJ4

4 0
2 years ago
Which of the following could be true of two species that have a competitve ralationship in the same ecosystems Help me out pleas
ankoles [38]
The answer is b. as a whole, the species is mutually beneficial to carry on each others traits and exist in the same ecosystem.
6 0
3 years ago
Read 2 more answers
i am begging anyone to help me with this! (all tutors i've asked said they can't solve it but i need someone to help me out) - i
9966 [12]

First, we need to calculate how much energy we will get from this combustion.

Assuming the combustion is complete, we have the octane reacting with O₂ to form only water and CO₂, so:

C_8H_{18}+O_2\to CO_2+H_2O

We need to balance the reaction. Carbon only appear on two parts, so, we can start by it:

C_8H_{18}+O_2\to8CO_2+H_2O

Now, we balance the hydrogen:

C_8H_{18}+O_2\to8CO_2+9H_2O

And in the end, the oxygen:

C_8H_{18}+\frac{25}{2}O_2\to8CO_2+9H_2O

We can multiply all coefficients by 2 to get integer ones:

2C_8H_{18}+25O_2\to16CO_2+18H_2O

Now, we need to use the enthalpies of formation to get the enthalpy of reaction of this reaction.

The enthalpy of reaction can be calculated by adding the enthalpies of formation of the products multiplied by their stoichiometric coefficients and substracting the sum of enthalpies of formation of the reactants multiplied by their stoichiometric coefficients.

For the reactants, we have (the enthalpy of formation of pure compounds is zero, which is the case for O₂):

\begin{gathered} \Delta H\mleft\lbrace reactants\mright\rbrace=2\cdot\Delta H\mleft\lbrace C_8H_{18}\mright\rbrace+25\cdot\Delta H\mleft\lbrace O_2\mright\rbrace \\ \Delta H\lbrace reactants\rbrace=2\cdot(-250.1kJ)+25\cdot0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ+0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ \end{gathered}

For the products, we have:

\begin{gathered} \Delta H_{}\mleft\lbrace product\mright\rbrace=16\cdot\Delta H\lbrace CO_2\rbrace+18\cdot\Delta H\lbrace H_2O\rbrace \\ \Delta H_{}\lbrace product\rbrace=16\cdot(-393.5kJ)+18\cdot(-285.5kJ) \\ \Delta H_{}\lbrace product\rbrace=-6296kJ-5139kJ \\ \Delta H_{}\lbrace product\rbrace=-11435kJ \end{gathered}

Now, we substract the rectants from the produtcs:

\begin{gathered} \Delta H_r=\Delta H_{}\lbrace product\rbrace-\Delta H\lbrace reactants\rbrace \\ \Delta H_r=-11435kJ-(-500.2kJ) \\ \Delta H_r=-10934.8kJ \end{gathered}

Now, this enthalpy of reaction is for 2 moles of C₈H₁₈, so for 1 mol of C₈H₁₈ we have half this value:

\Delta H_c=\frac{1}{2}\Delta H_r=\frac{1}{2}\cdot(-10934.8kJ)=-5467.4kJ

Now, we have 100 g of C₈H₁₈, and its molar weight is approximately 114.22852 g/mol, so the number of moles in 100 g of C₈H₁₈ is:

\begin{gathered} M_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{n_{C_8H_{18}}} \\ n_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{M_{C_8H_{18}}}=\frac{100g}{114.22852g/mol}\approx0.875438mol \end{gathered}

Since we have approximately 0.875438 mol, and 1 mol releases -5467.4kJ when combusted, we have:

Q=-5467.4kJ/mol\cdot0.875438mol\approx-4786.37kJ

Now, for the other part, we need to calculate how much heat it is necessary to melt a mass, <em>m</em>.

First, we have to heat the ice to 0 °C, so:

\begin{gathered} Q_1=m\cdot2.010J/g.\degree C\cdot(0-(-10))\degree C \\ Q_1=m\cdot2.010J/g\cdot10 \\ Q_1=m\cdot20.10J/g \end{gathered}

Then, we need to melt all this mass, so we use the latent heat now:

Q_2=n\cdot6.03kJ/mol

Converting mass to number of moles of water we have:

\begin{gathered} M=\frac{m}{n} \\ n=\frac{m}{M}=\frac{m}{18.01528g/mol} \end{gathered}

So:

Q_2=\frac{m}{18.01528g/mol}_{}\cdot6.03kJ/mol\approx m\cdot0.334716kJ/g

Adding them, we have a total heat of:

\begin{gathered} Q_T=m\cdot20.10J/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.02010kJ/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.354816kJ/g \end{gathered}

Since we have a heat of 4786.37 kJ form the combustion, we input that to get the mass (the negative sign is removed because it only means that the heat is released from the reaction, but now it is absorbed by the ice):

\begin{gathered} 4786.37kJ=m\cdot0.354816kJ/g \\ m=\frac{4786.37kJ}{0.354816kJ/g}\approx13489g\approx13.5\operatorname{kg} \end{gathered}

Since we have a total of 20kg of ice, we can clculate the percent using it:

P=\frac{13.5\operatorname{kg}}{20\operatorname{kg}}=0.675=67.5\%

5 0
1 year ago
if 30.0L of oxygen are cooled from 200 degrees celsius to 1 degree celsius at constant pressure, what is the new volume of oxyge
Sergio039 [100]

Answer: 18.65L

Explanation:

Given that,

Original volume of oxygen (V1) = 30.0L

Original temperature of oxygen (T1) = 200°C

[Convert temperature in Celsius to Kelvin by adding 273.

So, (200°C + 273 = 473K)]

New volume of oxygen V2 = ?

New temperature of oxygen T2 = 1°C

(1°C + 273 = 274K)

Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law

V1/T1 = V2/T2

30.0L/473K = V2/294K

To get the value of V2, cross multiply

30.0L x 294K = 473K x V2

8820L•K = 473K•V2

Divide both sides by 473K

8820L•K / 473K = 473K•V2/473K

18.65L = V2

Thus, the new volume of oxygen is 18.65 liters.

5 0
3 years ago
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