What is the resistance of a fluid on an object?

1 answer:

5 0

Fluid resistance refers to the forces a fluid places on a moving object in the opposite direction to the movement, also known as drag. Drag is the force created by a fluid to resist the motion of an object through it. There are 2 main sources of drag: pressure (form) and friction.

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Susan and her friend found a piece of metal which they want to

joja [24]

Answer:

The given metal is beryllium.

Explanation:

Given data:

Mass of metal = 74 g

Volume of metal = 40 cm³

Which metal is this = ?

Solution:

First of all we will calculate the density.

it is mass divided by volume.

d = m/v

d = 74 g/40 cm³

d = 1.85 g/cm³

In literature it is given that the density of beryllium is 1.85 g/cm³. Thus given metal is beryllium.

4 0

3 years ago

How many moles of water are In 44.99 grams

german

44.99g H2O x (1 mol H2O / 18.02g H2O) x (6.022x10^23 molecules H2O / 1 mol H2O) = 1.503x10^24 molecules H2O

6 0

3 years ago

What is the volume of 1.9 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)?

masya89 [10]

Considering ideal gas behavior, the volume of 1 mol of gas at STP is 22.4 L; then the volume occupied by 1.9 moles is 1.9mol*22.4L/mol = 42. 6 L.

Answer: 43 L

5 0

3 years ago

Read 2 more answers

3.65 gram of hcl is dissolved in 180 gram of water. Find the total number of molecules of hydrogen

Morgarella [4.7K]

Answer:

$Molec_{\ H_{tot}}=1.206x10^{25}molec$

Explanation:

Hello.

In this case, taking into account that HCl has one molecule of hydrogen per mole of compound which weights 36.45 g/mol, we compute the number of molecules of hydrogen in hydrochloric acid by considering the given mass and the Avogadro's number:

$molec_{\ H}=3.65gHCl*\frac{1molHCl}{36.45gHCl} *\frac{1molH}{1molHCl}*\frac{6.022x10^{23}molec_\ H}{1molH} =6.03x10^{22}molec$

Now, from the 180 g of water, we see two hydrogen molecules per molecule of water, thus, by also using the Avogadro's number we compute the molecules of hydrogen in water:

$molec_{\ H}=180gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O}*\frac{6.022x10^{23}molec_\ H}{1molH} =1.20x10^{25}molec$