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3 years ago

13

In the graph below, the acceleration of an object is plotted against the net force applied to the object. When a net force of 2

newtons is applied,
A. 1 m/s^2
B. 2 m/s^2
C. 3 m/s^2
D. 4 m/s^2

1 answer:

Westkost [7]3 years ago

3 0

Answer:

The answer to your question is: letter A.

Explanation:

In this conditions we observe that the acceleration is 1 m/s².

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What’s molar mass? For chemistry

Alik [6]

In chemistry, the molar mass of a chemical compound is defined as the mass of a sample of that compound divided by the amount of substance in that sample, measured in moles.

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3 years ago

A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s

Alla [95]

<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol$

The chemical equation for the reaction of 1-butanol and NaBr is:

$\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}$

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = $\frac{1}{1}\times 0.108=0.108$ moles of 1-bromobutane

Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

$0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g$

To calculate the percentage yield of 1-bromobutane, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

$\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%$

Hence, the percent yield of the 1-bromobutane is 48.65 %

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3 years ago

Si la eficiencia de una máquina de calor es del 50% significa que ?

Karolina [17]

Answer:

No toques el link

Explanation:

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3 years ago

4HCl(g)+O2(g)⟶2Cl2(g)+2H2O(g) Calculate the number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2. mass:

FromTheMoon [43]

The number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2 is 13.6675 g.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × $10^{23}$ of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Given data:

Moles of hydrochloric acid = 0.385 mol

Mass of chlorine gas =?

Chemical equation:

4HCl + O₂ → 2Cl₂ + 2H₂O

Now we will compare the moles of Cl₂ with HCl.

HCl : Cl₂

4 : 2

0.385 : 2÷4× 0.385 = 0.1925 mol

Oxygen is present in excess that's why the mass of chlorine produced depends upon the available amount of HCl.

Mass of Cl₂ :

Mass of Cl₂ = moles × molar mass

Mass of Cl₂ =0.1925 mol × 71 g/mol

Mass of Cl₂ = 13.6675 g

Hence, the number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2 is 13.6675 g.

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2 years ago

Which of the following is defined as the mass of an atom based on the mass of an atom of carbon-12?

iogann1982 [59]

<span>This would be the atomic mass. In an atom of carbon-12, there are 6 protons and 6 neutrons at rest (electrons have a negligible mass and are usually not part of the overall mass calculation). All atomic masses are based off the measurements of this specific iteration of carbon.</span>

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3 years ago

Read 2 more answers