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user100 [1]
3 years ago
10

2/5 in metres??..help please ​

Chemistry
1 answer:
liubo4ka [24]3 years ago
5 0

Answer:

0.0102

Explanation:

fist divide 2/5 which is 0.4 then you divide the value by 39.37 because one meter is 39.37 inches

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What factors affect the geometry of a molecule?
natali 33 [55]

The factors that affect geometry of a molecule are

> The number of bonding electron pairs around the central atom.

> The number of pairs of non-bonding ("lone pair") electrons around the central atom.

4 0
3 years ago
How many electrons are shared in the Lewis structure
Diano4ka-milaya [45]
Hey there,

Answer:

4 valence electrons.

Hope this helps :D

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3 0
3 years ago
COCI2 has an effusion rate of 0.00172 m/sec. Which of the gases below would have an effusion rate of 0.00323 m/sec?
geniusboy [140]
CO por qué si y punto, chao
8 0
2 years ago
Balance this equation. Pb(NO3)2(aq)+NaCl(aq) -&gt; NaNO3(aq)+PbCl2(s)
const2013 [10]

Answer:

Pb(NO3)2(aq) + 2NaCl(aq) -> 2NaNO3(aq)+PbCl2(s)

Explanation:

Pb(NO3)2(aq)+NaCl(aq) -> NaNO3(aq)+PbCl2(s)

This is how it starts out.

Left:

  • 2 NO3s
  • 1 Pb
  • 1 Na
  • 1 Cl

Right

  • 1 Na
  • 1 NO3
  • 1 Pb
  • 2 Cl

So the place to start with this equation is to bring the Cls up to 2

Pb(NO3)2(aq)+2NaCl(aq) -> NaNO3(aq)+PbCl2(s)

But the Nas are now out of kilter.

Pb(NO3)2(aq)+ 2NaCl(aq) -> NaNO3(aq)+PbCl2(s)

Now the right has a problem. There's only 1 Na

Pb(NO3)2(aq) + 2 NaCl(aq) -> 2NaNO3(aq)+PbCl2(s)

Check it out. It looks like we are done.

5 0
3 years ago
Use the following balanced reaction to solve:
Naily [24]

Answer:  60.7 g of PH_3 will be formed.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar volume}}    

\text{Moles of} H_2=\frac{60L}{22.4L}=2.68moles

The balanced chemical reaction is

P_4(s)+6H_2(g)\rightarrow 4PH_3(g)

H_2 is the limiting reagent as it limits the formation of product and P_4 is the excess reagent.

According to stoichiometry :

6 moles of H_2 produce = 4 moles of PH_3

Thus 2.68 moles of H_2 will produce=\frac{4}{6}\times 2.68=1.79moles  of PH_3

Mass of PH_3=moles\times {\text {Molar mass}}=1.79moles\times 33.9g/mol=60.7g

Thus 60.7 g of PH_3 will be formed by reactiong 60 L of hydrogen gas with an excess of P_4

3 0
3 years ago
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