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disa [49]
4 years ago
6

A positively charged particle is placed next to an negatively charged object. In which direction does the positively charged par

ticle move? Explain why.
Chemistry
1 answer:
Arte-miy333 [17]4 years ago
4 0

Answer:

The positively charged particle will go towards the negatively charged particle because opposites attract.

Explanation:

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Calculate the number of moles of H2 produced in the reaction of Mg(s) with HCl(aq). Mg(s) is the
Taya2010 [7]

Explanation:

Moles of metal,

=

4.86

⋅

g

24.305

⋅

g

⋅

m

o

l

−

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=

0.200

m

o

l

.

Moles of

H

C

l

=

100

⋅

c

m

−

3

×

2.00

⋅

m

o

l

⋅

d

m

−

3

=

0.200

m

o

l

Clearly, the acid is in deficiency ; i.e. it is the limiting reagent, because the equation above specifies that that 2 equiv of HCl are required for each equiv of metal.

So if

0.200

m

o

l

acid react, then (by the stoichiometry), 1/2 this quantity, i.e.

0.100

m

o

l

of dihydrogen will evolve.

So,

0.100

m

o

l

dihydrogen are evolved; this has a mass of

0.100

⋅

m

o

l

×

2.00

⋅

g

⋅

m

o

l

−

1

=

?

?

g

.

If 1 mol dihydrogen gas occupies

24.5

d

m

3

at room temperature and pressure, what will be the VOLUME of gas evolved?

5 0
2 years ago
What is the molar mass of Ba3(PO4)2?
Yakvenalex [24]

Answer:

<h3>601.93 g/mol</h3>

<h3>explanation:</h3>

Problem: The Ba3(PO4)2 (molar mass = 601.93 g/mol) precipitate that formed from a salt mixture has a mass of 0.667 g.

5 0
2 years ago
Lead(II) oxide from an ore can be reduced to elemental lead by heating in a furnace with carbon. Calculate the expected yield of
Lelu [443]

Answer:

53kg is the expected yield of lead

Explanation:

Firstly, in order to solve this question, we need to write the equation of reaction correctly. This is as follows:

PbO(s) + C(s) ---> Pb(l) + CO(g)

We proceed from here. We should get the limiting reactant but this can only be obtained by getting the number of moles of each reactant present.

The formula to use across all boards is that the number of moles is the mass of each of the reactant divided by the molar mass of each of the reactant.

For PBO, mass is 57kg = 57000g

Molar mass of PBO = 223.20g/mol

The number of moles is thus 57,000/223.2 = 255.37 moles

For carbon, mass is also 57kg = 57000g

Molar mass is 12g/mol

Number of moles of carbon = 57000/12 = 4750 moles

From the number of moles, we can see that the number of moles of Carbon is greater than that of PbO. This means that PbO is the limiting reagent.

Hence we use it to calculate percentage yield.

The number of moles of lead formed is the same of number of moles of lead oxide = 255.37 since we have mole ratio of 1 to 1

The molar mass of lead is 207.20g/mol

The mass of lead formed is = moles of lead formed * molar mass of lead = 207.20 * 255.37 = 52,912g which is approximately 53kg

Hence the expected yield is 53kg

6 0
3 years ago
What are the base units the SI units are based on?<br> A-10<br> B-100<br> C-1,000<br> D-1
Olegator [25]

Answer:

A-10

Explanation:

In the SI, designations of multiples and subdivision of any unit may be arrived at by combining with the name of the unit the prefixes deka, hecto, and kilo meaning, respectively, 10, 100, and 1000, and deci, centi, and milli, meaning, respectively, one-tenth, one-hundredth, and one-thousandth.

IM NOT SURE PO

4 0
2 years ago
Un tanque de acetileno para una antorcha de soldadura de oxiacetileno proporciona 9340 L de gas acetileno, C2H2, a 0°C y 1 atm 2
Helen [10]

Answer:

3.33 tanques de O₂

Explanation:

Basados en la reacción:

2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(g)

<em>2 moles de acetileno reaccionan con 5 moles de oxígeno produciendo 4 moles de dióxido de carbono y 2 moles de agua</em>

<em />

La ley de Avogadro dice que el volumen de un gas bajo temperatura y presión constantes es proporcional a las moles de este gas. Así, como 2 moles de acetileno reaccionan con 5 moles de oxígeno, los litros de O₂ necesarios para quemar 9340L de acetileno son:

9340 L C₂H₂ × (5 moles O₂ / 2 moles C₂H₂) = <em>23350L de O₂</em>

Si un tanque contiene 7x10³ L de O₂ serán necesarios:

23350L O₂ ₓ (1 tanque / 7x10³L) =<em> 3.33 tanques de O₂</em>

6 0
3 years ago
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