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stiks02 [169]
3 years ago
13

0.0055 in scientific notation

Chemistry
2 answers:
Digiron [165]3 years ago
8 0

Answer:

5.5 x 10^-3

Explanation:

user100 [1]3 years ago
4 0

Answer:

5.5 × 10^-3

Explanation:

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A steel cylinder has a pressure of 550 torr at 177 degrees Celsius. What will be the pressure if the temperature is decreased by
Schach [20]

Answer:

P_2=507.7torr

Explanation:

Hello there!

In this case, when we have a gas that is undergoing a change in both pressure and temperature, we utilize the Gay-Lussac's equation as shown below:

\frac{T_2}{P_2}=\frac{T_1}{P_1}

Thus, since we are given the initial pressure and temperature and the final temperature, we can compute the final pressure as shown below:

P_2=\frac{P_1}{T_1}*T_2

So we plug in, by making sure the temperatures are in kelvins, to obtain:

P_2=\frac{550torr}{(117+273)K}*(87+273)K\\\\P_2=507.7torr

Best regards!

8 0
3 years ago
Enter the formula for the polyatomic ion in MnCO3.
joja [24]
The formula is Mn2+CO2= MnCO3
8 0
3 years ago
When a pendulum swings, at which point is kinetic<br> energy highest:
Anna [14]

Answer:

3

Explanation:

3 0
3 years ago
Concentrated hydrochloric acid is an aqueous solution that is 34.70 % HCl. The density of the solution is 1.19 g/mL. What mass o
mixas84 [53]

The mass of HCl that is contained in the solution is 147 g HCl

Why?

To find the mass of HCl we have to apply what is called a conversion factor. In a conversion factor we put the units we don't want at the bottom, and the ones we want at the top.

For this question, we want to go from liters of solution to mass of HCl, and the conversion factor is laid out as follows:

0.356Lsolution*\frac{1000mL}{1L}*\frac{1.19 g solution}{1 mL solution}*\frac{34.70 g HCl}{100 g solution}=147 g HCl

Have a nice day!

#LearnwithBrainly

3 0
4 years ago
If 34.5 g of Copper reacts with 70.2 g of silver nitrate, according to the following
Vlada [557]

Answer:

44,55 can be produced.

Explanation:

First, we balanced the equation

1Cu + 2AgNO3 → 1Cu(NO3)2 + Ag

Then, we find the moles of each reagent

mol Cu = \frac{34,5g}{63,55\frac{g}{mol} } = 0,543 mol\\

mol AgNO3 = \frac{70,2g}{169,87\frac{g}{mol} } = 0,413 mol\\

Now, we find the limiting reagent from the quantities of product that can be formed from each reagent

mol AgNO3= 0,543mol Cu . \frac{2 mol AgNo3}{1 mol Cu} = 1,086 mol

mol Cu= 0,413 mol AgNO3 . \frac{1 mol Cu}{2 mol AgNO3} = 0,206 mol

1,086 moles of AgNO3 is necessary for each mole of Cu since we have 0.413 moles of Ag(NO3), the nitrate is the limiting reagent

the value of the limiting reagent determines the amount of product that is generated

∴ 0,413 mol of Ag can be produced

Ag = 0,413 mol . 107,87\frac{g}{mol} = 44,55g

Ag≈ 44,6g

4 0
4 years ago
Read 2 more answers
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