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2 years ago

Cuales son las fórmulas de la velocidad

Chemistry

1 answer:

AnnZ [28]2 years ago

7 0

Answer:

s = d÷ t

Explanation:

Where s means speed, d means distance and t means time

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Calculate the amount of mole of gas which occupies 250cm at S.T.P

solmaris [256]

it is already given

brainly.com/question/17158435

3 0

2 years ago

A rectangular field measures 6.0 m by 8.0 m. What is the area of the field in square centimeters (cm2)? Use the formula: Area =

zhenek [66]

The correct answer is 4800 cm^2

Step By Step Answer:
Length times width:
6.0 * 8.0 = 48
Translate m to cm:
4800 cm^2

Help this helps!

7 0

2 years ago

Read 2 more answers

A 115.0-g sample of oxygen was produced by heating 400.0 g of potassium chlorate.2KClO3 Right arrow. 2KCI + 3O2What is the perce

STALIN [3.7K]

Answer:

73.4% is the percent yield

Explanation:

2KClO₃ → 2KCl + 3O₂

This is a decomposition reaction, where 2 moles of potassium chlorate decompose to 2 moles of potassium chloride and 3 moles of oxygen.

We determine the moles of salt: 400 g . 1. mol /122.5g= 3.26 moles of KClO₃

In the theoretical yield of the reaction we say:

2 moles of potassium chlorate can produce 3 moles of oxygen

Therefore, 3.26 moles of salt, may produce (3.26 . 3) /2 = 4.89 moles of O₂

The mass of produced oxygen is: 4.89 mol . 32 g /1mol = 156.6g

But, we have produced 115 g. Let's determine the percent yield of reaction

Percent yield = (Produced yield/Theoretical yield) . 100

(115g / 156.6g) . 100 = 73.4 %

5 0

2 years ago

620 mL of nitrogen at standard pressure is compressed into a 480 mL container. What is the new pressure in kPa?

bulgar [2K]

Answer:

c. 131 kPa

Explanation:

Hello!

In this case, since the relationship between volume and pressure is inversely proportional, based on the Boyle's law:

$P_1V_1=P_2V_2$

Considering that the standard pressure is 101.325 kPa, we can compute the final pressure as shown below:

$P_2=\frac{P_1V_1}{V_2}=\frac{620mL*101.325kPa}{480mL}\\\\P_2=131kPa$

Therefore, the answer is c. 131 kPa .

Best regards!!

7 0

2 years ago

Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)

Dovator [93]

Answer:

1. 0.97 V

2. $Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)$

Explanation:

In this case, we can start with the half-reactions:

$Ag^+~_(_a_q_)->~Ag_(_s_)$

$Al_(_s_)~->~Al^+^3~_(_a_q_)$

With this in mind we can add the electrons:

$Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)$ Reduction

$Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~$ Oxidation

The reduction potential values for each half-reaction are:

$Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_)$ - 0.69 V

$Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_)$ -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to flip the reduction potential value:

$Al_(_s_)~->~Al^+^3~+~2e^-~$ +1.66 V

Finally, to calculate the overall potential we have to add the two values:

1.66 V - 0.69 V = 0.97 V

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

$Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)$

I hope it helps!

3 0

3 years ago