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kotegsom [21]
2 years ago
12

An experiment requires that each student use an 8.5 cm length of magnesium ribbon. How many students can do the experiment if th

ere is a 570 cm length of magnesium ribbon available?
Chemistry
2 answers:
melisa1 [442]2 years ago
7 0
570/8.5=67.0 58... you only have to take the natural part, si the answer is 67 students
BlackZzzverrR [31]2 years ago
5 0

In the particular experiment, each student uses 8.5 cm of ribbon.

The total length of the Mg ribbon present is 570 cm

so to find how many students can use the ribbon

number of students = total length of the ribbon / length each student uses

number of students = 570 cm / 8.5 cm/student = 67. 1 students

since it should be a whole number it should be 67

therefore 67 students can use the ribbon

You might be interested in
2C6H5COOH + 15O2 → 14CO2 + 6H2O which of the following options gives the correct product:product ratio?
mash [69]
You did not include the options but I can tell you the product ratio.

The product ratio is the mole ratio of the products of the reaction.

From the balanced chemical equation you have all the mole ratios:

The given equation is: 2 C6H5COOH + 15O2 --> 14 CO2 + 6H2O

The mole ratios are: 2 C6H5COOH: 15 O2: 14 CO2 : 6 H2O

The products are CO2 and H2O

Their mole ratio = 14 CO2 : 6 H2O

That can be expressed as:

14 mol CO2        7 mol CO2
----------------- =  -----------------
  6 mol H2O        3 mol H2O

It is also the same that:

6 mol H2O : 14 mol CO2

  6 mol H2O           3 mol H2O
------------------ =  -------------------
14 mol CO2           7 mol CO2

So, compare your options to the ratios show above and pick the proper ratio.
7 0
2 years ago
water’s molar mass is 18.01 g/mol. The molar mass of glycerol is 92.09 g/mol. At 25 celsius, glycerol is more viscous than water
Gemiola [76]

Answer is: glycerol because it is more viscous and has a larger molar mass.

Viscosity depends on intermolecular interactions.

The predominant intermolecular force in water and glycerol is hydrogen bonding.

Hydrogen bond is an electrostatic attraction between two polar groups in which one group has hydrogen atom (H) and another group has highly electronegative atom such as nitrogen (like in this molecule), oxygen (O) or fluorine (F).

4 0
2 years ago
Heavy nuclides with too few neutrons to be in the band of stability are most likely to decay by what mode?
Margarita [4]

Answer:

Positron emission

Explanation:

Positron emission involves the conversion of a proton to a neutron. This process increases the mass number of the daughter nucleus by 1 while its atomic number remains the same. The new neutron increases the number of neutrons present in the daughter nucleus hence the process increases the N/P ratio.

A positron is usually ejected in the process together with an anti-neutrino to balance the spins.

6 0
1 year ago
A 75 lb (34 kg) boy falls out of a tree from a height of 10 ft (3 m). i. What is the kinetic energy of the boy when he hits the
Jobisdone [24]

Answer:

Kinetic energy of boy just before hitting the ground is \approx1000 J.

Speed of boy just before hitting the ground is 7.67 m/s

or 17.16 mi/hr.

Explanation:

Given that:

Mass of boy = 75lb = 34 kg

Height, h = 10ft = 3m

To find:

Kinetic energy of boy when he hits the ground.

<em>As per law of conservation of energy The potential energy gets converted to kinetic energy.</em>

<em></em>

<em></em>\therefore<em> </em>Kinetic energy at the time boy hits the ground = Initial potential energy of the boy when he was at the Height 'h'

The formula for potential energy is given as:

PE = mgh

Where m is the mass

g is the acceleration due to gravity, g = 9.8 m/s^2

h is the height of object

Putting all the values:

PE = 34 \times 9.8 \times 3 \approx 1000\ J

Hence, Kinetic energy is \approx1000 J.

Formula for Kinetic energy is:

KE = \dfrac{1}{2}mv^2

where m is the mass and

v is the speed

Putting the values and finding v:

1000 = \dfrac{1}{2}\times 34 \times v^2\\\Rightarrow v^2 = 58.82\\\Rightarrow v = 7.67\ m/s

Given that:

1 m = 1.094 yd and 1 mi = 1760 yd

\Rightarrow 1609\ m = 1\ mi

Converting 7.67 m/s to miles/hour:

\dfrac{7.67 \times 3600}{1609}=17.16\ mi/h

4 0
1 year ago
In the reaction C + O2 → CO2, 18 g of carbon react with oxygen to produce 72 g of carbon dioxide. What mass of oxygen would be n
bulgar [2K]
<span>Molar mass(C)= 12.0 g/mol
Molar mass (O2)=2*16.0=32.0 g/mol
Molar mass (CO2)=44.0 g/mol

18g C*1mol C/12 g C = 1.5 mol C

                                 C +     O2 →                CO2

from reaction       1 mol    1 mol              1 mol
from problem     1.5 mol   1.5 mol         1.5 mol

1.5 mol O2*32 g O2/1 mol O2 = 48 g O2

In reality this reaction requires only 48 g O2 for 18 g carbon.
And from 18 g carbon you can get only
1.5 mol CO2*44 g CO2/1 mol CO2=66 g CO2
But these problem has 72g CO2. The best that we can think, it is a mix of CO2 and O2.
So to find all amount  of O2  that was added for the reaction (probably people who wrote this problem wanted this)
we need  (the mix of 72g - mass of carbon 18 g)= 54 g.
So the only answer that is possible is 
</span><span>2.) 54 g.</span>
3 0
2 years ago
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