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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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6c.Calculate the maximum volume, in dm3, of chlorine gas at Stp that can be obtained from 23.4 tonnes of molten sodium chloride.

sweet [91]

Answer:

4.48×10⁶ dm³

Explanation:

We'll begin by converting 23.4 tonnes to grams (g). This can be obtained as follow:

1 tonne = 10⁶ g

Therefore,

23.4 tonnes = 23.4 × 10⁶

23.4 tonnes = 2.34×10⁷ g

Thus, 23.4 tonnes is equivalent to 2.34×10⁷ g

Next, we shall determine the number of mole in 2.34×10⁷ g of NaCl. This can be obtained as follow:

Mass NaCl = 2.34×10⁷ g

Molar mass of NaCl = 58.5 g/mol

Mole of NaCl =?

Mole = mass / molar mass

Mole of NaCl = 2.34×10⁷ / 58.5

Mole of NaCl = 4×10⁵ moles

Next, we shall determine the number of mole of chlorine, Cl₂ produced from the reaction. This can be obtained as follow:

2NaCl —> 2Na + Cl₂

From the balanced equation above,

2 moles of NaCl reacted to produce 1 mole of Cl₂.

Therefore, 4×10⁵ moles of NaCl will react to produce = (4×10⁵ × 1)/2 = 2×10⁵ moles of Cl₂.

Thus, 2×10⁵ moles of Cl₂ were obtained from the reaction.

Finally, we shall determine the volume of Cl₂ produced. This can be obtained as follow:

1 mole of Cl₂ at stp = 22.4 dm³

Therefore,

2×10⁵ moles of Cl₂ at stp = 2×10⁵ 22.4

2×10⁵ moles of Cl₂ at stp = 4.48×10⁶ dm³

Thus, the volume of chlorine obtained from the reaction is 4.48×10⁶ dm³

6 0

3 years ago

Solve: Turn off Show summary. Use the Choose reaction drop down menu to see other equations, and balance them. Check your answer

suter [353]

Answer: See below

Explanation:

To balance equations, you want to have the same amount of elements on the product and reactants side.

__Al+ __HCl→__AlCl₃+ __H₂

We see that there are 3 Cl on the products side and 1 on the reactants side, but there are 2 H on the product and 1 on reactant. To fulfill them both, let's put a 6 at HCl.

__Al+ 6HCl→__AlCl₃+ __H₂

Now that we have a 6 at HCl, we can fill in AlCl₃ and H₂.

__Al+ 6HCl→ 2AlCl₃+ 3H₂

All we have left is to fill in Al.

2Al+ 6HCl→ 2AlCl₃+ 3H₂

-----------------------------------------------------------------------------------------------------------------

__NaCl→ __Na+ __Cl₂

Since we have 2 Cl on the products, we must put 2 on the reactants.

2NaCl→ __Na+ 1Cl₂

With 2 NaCl, we can fill in Na.

2NaCl→ 2Na+ 1Cl₂

-----------------------------------------------------------------------------------------------------------------

__Na₂S+ __HCl→ __NaCl+ __H₂S

We see 2 Na on reactants, so we can put 2 on the products.

__Na₂S+ __HCl→ 2NaCl+ __H₂S

With 2 H and 2 Cl on the products, we can put a 2 at HCl.

1Na₂S+ 2HCl→ 2NaCl+ 1H₂S

7 0

3 years ago

a new weight-loss fad claims that you can reduce your mass simply by traveling to the Moon where gravity is weaker. why is this

Lilit [14]

Your mass can never change no matter where you are since mass is the amount of matter you contain. By going to the moon, you do become lighter due to the weaker gravity but the amount of matter that you are made of (your mass) does not change.

I hope the helps. Let me know if anything is unclear.

3 0

3 years ago

(a) How many stereoisomers are possible for 4,4-dimethyl-1,2-cyclopentanediol?

Monica [59]

Answer:

(a) 4 stereoisomers: (1R,2S) 4,4-dimethylcyclopentane1,2-diol; (1R,2S) 4,4-dimethylcyclopentane1,2-diol; (1S,2S) 4,4-dimethylcyclopentane1,2-diol; (1R,2R) 4,4-dimethylcyclopentane1,2-diol. (b) (1R,2S) 4,4-dimethylcyclopentane1,2-diol; (1R,2S) 4,4-dimethylcyclopentane1,2-diol (c) No, they are optically inactive.

Explanation:

(a) The stereoisomers possible for 4,4-dimethyl-1, 2-cyclopentanediol are (1R,2S) 4,4-dimethylcyclopentane1,2-diol; (1R,2S) 4,4-dimethylcyclopentane1,2-diol; (1S,2S) 4,4-dimethylcyclopentane1,2-diol; (1R,2R) 4,4-dimethylcyclopentane1,2-diol. Their structures are shown in the attached file.

(b) The oxidation reaction that occurs between 4,4-dimethylcyclopentene and osmium tetroxide produces (1R,2S) 4,4-dimethylcyclopentane1,2-diol; (1R,2S) 4,4-dimethylcyclopentane1,2-diol. Their structures are also shown in the attached file.

(c) The products of the oxidation reaction in (b) are optically inactive because the compounds contain plane of symmetry that makes them optically inactive compound.

5 0

4 years ago

There is some gas in a container at pressure 136 atm. When 142 dm3 gas is released at pressure of 1 atm,then the pressure of the

mamaluj [8]

Answer:

228.7 $dm^{3}$

Explanation:

According to Boyle's law

P1V1 = P2V2

where P1& P2 are initial and final pressure

and V1& V2 and initial and final volumes respectively.

Given:

P1 = 136atm, P2 = 131.6atm

V1 = 142 $dm^{3}$ ; V2 = ?

V2 = $\frac{P1V1}{P2}$ = $\frac{136*142}{131.6}$

= 146.7 $dm^{3}$

Total volume, V = V1 + V2

=146.7 + 142

<u>= 228.7</u> $dm^{3}$ <u></u>

8 0

3 years ago