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Goshia [24]
3 years ago
5

A force of 100 newtons is applied to a box at an angle of 36º with the horizontal. If the mass of the box is 25 kilograms, what

is the horizontal acceleration of the box? 1.52 meters/second2
Chemistry
1 answer:
notka56 [123]3 years ago
5 0

Answer:

a_x=4.944m/s^2

Explanation:

Hello,

Based on the acting force that is applied horizontally, one can propose the following equation based on Newton's laws:

F=ma

Nevertheless, since we've got a angled force, it becomes:

F=macos\theta

In this case, cos\theta accounts for the formed angle, so the horizontal acceleration turns out into:

a_x=\frac{F}{mcos\theta}=\frac{100kg*\frac{m}{s^2} }{25kgcos(36^0)}=4.944m/s^2

Best regads.

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20 poll
aleksley [76]

Answer: n=15.56moles

Explanation:

PV = nRT

where  

P is pressure in atmospheres

V is volume in Liters

n is the number of moles of the gas

R is the ideal gas constant = given as (0.0821L -atm/k-mol

PV = nRT

n= PV/RT

n= (1.5 X 230)/ (0.0821  X 270)

n= 15.56 moles

5 0
3 years ago
• We obtained the above 10.00-mL solution by diluting a stock solution using a 1.00-mL aliquot and placing it into a 25.00-mL vo
garik1379 [7]

Answer:

a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) 0.0035 mole

c)  0.166 M

Explanation:

Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.

The equation of the reaction is expressed as:

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

1 mole         3 mole

The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b)  if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

10 ml            17.50 ml

(x) M              0.200 M

Molarity = \frac{0.2*17.5}{1000}

= 0.0035 mole

c) What was the molar concentration of phosphoric acid in the original stock solution?

By stoichiometry, converting moles of NaOH to H₃PO₄; we have

= 0.0035 \ mole \ of NaOH* \frac{1 mole of H_3PO_4}{3 \ mole \ of \ NaOH}

= 0.00166 mole of H₃PO₄

Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:

Molar Concentration =  \frac{mole \ \ of \ soulte }{ Volume \ of \ solution }

Molar Concentration = \frac{0.00166 \ mole \ of \  H_3PO_4 }{10}*1000

Molar Concentration = 0.166 M

∴  the molar concentration of phosphoric acid in the original stock solution = 0.166 M

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3 years ago
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lara [203]
Volume= mass divided by density
V= m/d
55/3.23
= 17.03
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The addition of a catalyst to a chemical reaction provides an alternate pathway that
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