All categories

3 years ago

Why is oiling done time and again in a sewing machine

Chemistry

2 answers:

AnnZ [28]3 years ago

8 0

Answer:

Oiling is done on a regular basis to sewing Machine because in the machine there are lots of machinery parts i.e. wheels and axels, etc. which are in continuous use and they possess so much frictional force between them that wear tear may be observed in those parts.

Llana [10]3 years ago

7 0

For Frictionless sewing

You might be interested in

What is a covalent bond?

Sphinxa [80]

It is the electron sharing.
electronegative element + electronegative element

exemple :

O₂ , H₂

hope this helps!

3 0

3 years ago

Caption for this picture asap.

miss Akunina [59]

I will walk through different captions if one I say isn’t correct.
1) As the humans impact the environment, there will be consequences such as global warming. 2) When the long-term consequences has come, human life would become more difficult and complicated.
3) If the himan life hasn’t taken care of the global, it wouldn’t be safe and consequences would attracted by this cause. (As said above.)
Thank you!

3 0

3 years ago

How many grams are in 32.2 L of CO2?

Naddika [18.5K]

Answer:

63.25 grams of CO₂

Explanation:

To convert from liters to grams, we first need to convert from liters to moles. To do this, we divide the liters by 22.4, the amount of liters of a gas per mole.

32.2 / 22.4

= 1.4375 moles of CO₂

Now we want to convert from moles to grams. To do this, we multiply the moles by the molar mass of CO₂. The total molar mass can be found on the periodic table by adding up the molar mass of carbon (12) and two oxygen (32).

12 + 32 = 44

Now we want to multiply the moles by the molar mass.

1.4375 • 44

= 63.25 grams of CO₂

This is your answer.

Hope this helps!

7 0

3 years ago

10.00 mL of the final acid solution is reacted with excess barium chloride to produce a precipitate of barium sulfate (Fw: 233.4

Verizon [17]

Answer:- Actual molarity of the original sulfuric acid solution is 17.0M.

Solution:- Barium chloride reacts with sulfuric acid to make a precipitate of barium sulfate. The balanced equation is written as:

$BaCl_2(aq)+H_2SO_4(aq)\rightarrow BaSO_4(s)+2HCl(aq)$

From this equation there is 1:1 mol ratio between barium sulfate and sulfuric acid. So, if excess of barium chloride is added to sulfuric acid then the moles of sulfuric acid would be equivalent to the moles of barium sulfate. Moles of barium sulfate could be calculated from the mass of it's dry precipitate.

Molar mass of barium sulfate is 233.4 grams per mol. The calculations for the moles of sulfuric acid are given below:

$0.397gBaSO_4(\frac{1molBaSO_4}{233.4gBaSO_4})(\frac{1moH_2SO_4}{1molBaSO_4})$

= $0.00170molH_2SO_4$

From given information, 10.00 mL of final acid solution were taken to react with excess of barium chloride. It means 0.00170 moles of sulfuric acid are present in 10.0 mL of final acid solution. We could calculate the actual molarity of the final solution from here as:

10.0 mL = 0.0100 L

$molarity=\frac{0.00170mol}{0.0100L}$

= 0.170M

Now we would use the dilution equation to calculate the actual molarity of the original sulfuric acid solution. The molarity equation is:

$M_1V_1=M_2V_2$

From given information, 10.0 mL of original acid solution were taken in a 100 mL flask and water was added up to the mark. It means the 10 fold dilution is done. 10 fold dilution means the molarity becomes one tenth of it's original value. Let's do the calculations in reverse way as we have calculated the molarity of the final solution.

let's say the molarity after first dilution is Y. the volume is taken as 10.0 mL. Final volume is 100 mL and the molarity is 0.170M. Let's plug in the values in the equation:

Y(10.0mL) = 0.170M(100mL)

$Y=\frac{0.170M*100mL}{10.0mL}Y = 1.70MLet's do the similar calculations to find out the actual molarity of the original acid solution. Let's say the molarity of the original acid solution is X. 10.0 mL of it were taken and diluted to 100 mL on adding water. The molarity is 1.70M as is calculated in the above step. Let's plug in the values in the molarity equation again to solve it for X as:X(10.0mL) = 1.70M(100mL)[tex]X=\frac{1.70M*100mL}{10.0mL}$

X = 17.0M

Hence, the actual molarity of sulfuric acid solution is 17.0M.

5 0

3 years ago

A 3.664 g sample of a monoprotic acid was dissolved in water. It took 20.27 mL of a 0.1578 M NaOH solution to neutralize the aci

chubhunter [2.5K]

Answer:

Molar mass of monoprotic acid is 1145g/mol.

Explanation:

The reaction of a monoprotic acid (HX) with NaOH is:

HX + NaOH → H₂O + NaX

That means 1 mole of acid reacts with 1 mole of NaOH

If the neutralization of the acid spent 20.27mL of a 0.1578 M NaOH solution. Moles of NaOH are:

0.02027L × (0.1578mol / L) = <em>3.199x10⁻³ moles</em> NaOH ≡ moles HX.

As mass of the sample is 3.664g, molar mass of the acid is:

3.664g / 3.199x10⁻³ moles =<em> 1145g/mol</em>

5 0

4 years ago

Why is oiling done time and again in a sewing machine​

Why is oiling done time and again in a sewing machine