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iris [78.8K]
2 years ago
6

Two forces are applied to a 17kg box, as shown. The box is on a smooth surface

Chemistry
1 answer:
HACTEHA [7]2 years ago
5 0

Answer:

A. the box accelerated to at 1.0 m/s² to the right because the net force is 17N to the right.

Explanation:

To solve this problem, we need to find the net force acting on the body;

   Net force  = Forward force  - Backward force

  Forward force  = 35N

  Backward force  = 18N

 Net force  = 35N  - 18N = 17N

 So;

   Net force  = mass x acceleration

        17  = 17 x acceleration

        Acceleration  = 1m/s²

Therefore, the body moves in the forward direction with an acceleration of  1m/s²

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Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the fi
Savatey [412]

The question is incomplete, the complete question is;

Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the first heating, your crucible and contents weighs 17.51 g. After the second heating, your crucible and contents weighs 17.50 g.

What is the theoretical yield of sodium carbonate?

What is the experimental yield of sodium carbonate?

What is the percent yield for sodium carbonate?

Which errors could cause your percent yield to be falsely high, or even over 100%?

Answer:

See Explanation

Explanation:

We have to note that water is driven away after the second heating hence we are concerned with the weight of the pure dry product.

Hence;

From the reaction;

2 NaHCO3 → Na2CO3(s) + H2O(l) + CO2(g)

Number of moles of  sodium bicarbonate = 18.56 - 15.98 = 2.58 g/87 g/mol

= 0.0297 moles

2 moles of sodium bicarbonate yields 1 mole of sodium carbonate

0.0297 moles of 0.015 moles  sodium bicarbonate yields 0.0297 * 1/2 = 0.015 moles

Theoretical yield of sodium carbonate = 0.015 moles * 106 g/mol = 1.59 g

Experimental yield of sodium bicarbonate = 17.50 g - 15.98 g = 1.52 g

% yield = experimental yield/Theoretical yield * 100

% yield = 1.52/1.59 * 100

% yield = 96%

The percent yield may exceed 100% if the water and CO2 are not removed from the system by heating the solid product to a constant mass.

5 0
2 years ago
What is the concentration of H3O+ ions in saliva if [OH-] = 4.22 x 10-10 M? Provide the pH and the classification of this sample
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pH=4.625

The classification of this sample of saliva : acid

<h3>Further explanation</h3>

The water equilibrium constant (Kw) is the product of concentration

the ions:

Kw = [H₃O⁺] [OH⁻]

Kw value at 25° C = 10⁻¹⁴

It is known [OH-] =  4.22 x 10⁻¹⁰ M

then the concentration of H₃O⁺:

\tt 10^{-14}=4.22\times 10^{-10}\times [H_3O^+]\\\\(H_3O^+]=\dfrac{10^{-14}}{4.22\times 10^{-10}}=2.37\times 10^{-5}

pH=-log[H₃O⁺]\tt pH=5-log~2.37=4.625

Saliva⇒acid(pH<7)

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Svetradugi [14.3K]

Answer:

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