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True [87]
3 years ago
7

Neutrons have. a.negative b.positive c.no charge d.nucleus

Chemistry
1 answer:
Firdavs [7]3 years ago
5 0
A negative That's the answer
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Why do experiments need to be repeatable?
maw [93]
If they were not repeatable people would think the experiment is not accurate. If it can be repeated than the data can prove a very valid point. 
3 0
2 years ago
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For this reaction, C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O, the ∆H is –2200 kJ. If two moles of C3H8(g) reacted with excess oxygen,
algol13
<span> 1C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O, the ∆H is –2200 kJ

1mol </span>C3H8(g) ---- –2200 kJ
2mol C3H8(g)  ----2(–2200 kJ)=-4400kJ
4 0
3 years ago
A 20.0-milliliter sample of 0.200 M K2CO3 solution is added to 30.0 milliliters of 0.400 M Ba(NO3)2 solution.
jenyasd209 [6]

Answer:

(B) 0.160 M

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For K_2CO_3 :

Molarity = 0.200 M

Volume = 20.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.0×10⁻³ L

Thus, moles of K_2CO_3 :

Moles=0.200 \times {20.0\times 10^{-3}}\ moles

Moles of K_2CO_3 = 0.004 moles

For Ba(NO_3)_2 :

Molarity = 0.400 M

Volume = 30.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume =30.0×10⁻³ L

Thus, moles of Ba(NO_3)_2 :

Moles=0.400\times {30.0\times 10^{-3}}\ moles

Moles of Ba(NO_3)_2  = 0.012 moles

According to the given reaction:

K_2CO_3_{(aq)}+Ba(NO_3)_2_{(aq)}\rightarrow BaCO_3_{(s)}+2KNO_3_{(aq)}

1 mole of potassium carbonate react with 1 mole of barium nitrate

0.004 moles potassium carbonate react with 0.004 mole of barium nitrate

Moles of barium nitrate  = 0.004 moles

Available moles of barium nitrate  =  0.012 moles

Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of potassium carbonate gives 1 mole of barium carbonate

Also,

0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate

Mole of barium carbonate = 0.004 moles

Also, consumed barium nitrate = 0.004 moles  (barium ions precipitate with carbonate ions)

Left over moles = 0.012 - 0.004 moles = 0.008 moles

Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L

So, Concentration = 0.008/0.05 M = 0.160 M

<u>(B) is correct.</u>

4 0
3 years ago
PLSSS HEALP ASAP!!!! WILL REWARD
Liula [17]

A)1.75×3 moles of carbon monoxide

B)2:3

A)each mole of ferric oxide requires 3 moles of carbon monoxide. Therefore 1.75 moles requires 1.75 ×3 moles of carbon monoxide

6 0
1 year ago
Which condition increases the number of collisions between reactant molecules in a given volume?
Nezavi [6.7K]

Answer:

Increasing Surface Area

Explanation:

A greater surface area (meaning more, smaller particles) allows for more opportunity for particles to collide.  On the other hand, decreasing temperature and removing a catalyst would only decrease the number of collisions, and the clumping option doesn't make much sense.  Hope this helps!

4 0
2 years ago
Read 2 more answers
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