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3 years ago

How many grams of chlorine must be added to a balloon containing 8.00 grams of helium to double its volume?

1 answer:

4 0

The mole number of helium is 8/4=2 mol. Under same condition, the volume of gas is related to the mole number of the gas. So the mole number of chlorine is also 2 mol which is 142 g.

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Complete the table below by deciding whether a precipitate forms when aqueous solutions A and B are mixed. If a precipitate will

stiks02 [169]

The reaction between copper II chloride and sodium sulfide as well as lead II nitrate and potassium sulfate both produce precipitates.

The solubility of a substance in water is in accordance with the solubility rules. It is possible that a solid product may be formed when two aqueous solutions are mixed together. That solid product is referred to as a precipitate.

Now, we will consider each reaction individually to decode whether or not a precipitate is possible.

In the first reaction, we have; CuCl2(aq) + Na2S(aq) ---->CuS(s) + 2NaCl(aq). A precipitate (CuS) is formed.
In the second reaction, Pb(NO3)2(aq) + 2KNO3(aq) ----> PbSO4(s) + KNO3(aq), a precipitate PbSO4 is formed
In the third reaction, NH4Br(aq) + NaOH(aq) ----->NH3(g) + NaBr(aq) + H2O(l), a precipitate is not formed here.

Learn more: brainly.com/question/11969651

6 0

2 years ago

What is the maximum amount of HCl, in grams, that can be produced if 37.5 g of BCl3 and 60.0 g of H2O are reacted according to t

Rus_ich [418]

Ooooh boy alright. So, this may or may not be a limited reactant problem so we need to first find out of it is.

First, how many moles of each substance are there

the molar mass of BCl3 is <span>117.17 grams so 37.5 g / 117.17 is ~ .32 mol.
The molar mass of H2O is 18.02 so 60 / 18.02 is ~ 3.33 mol.

Now, for every 1 mole of BCl3, there are 3 moles of HCl created. Therefore, BCl3 can create ~ .96 moles.
For every 3 moles of H2O, there are 3 moles of HCl created. Therefore, HCl can create ~3.33 moles.

But, there is not enough BCl3 to support that 3.33 moles, only enough for .96 moles, therefore BCl3 is the limiting reactant. Now, to answer the question, simply multiply .96 moles by the molar mass of HCl.

.96 x 36.46 = ~35 g</span>

6 0

3 years ago

A bike rider approaches a hill with a speed of 8.5 m/s. The total mass of the bike and the rider is 85 kg. Find the kinetic ener

krek1111 [17]

KE=3070.625 J

Height = 3.686 m

<h3>Further explanation</h3>

mass of bike+rider=85 kg

velocity = 8.5 m/s

$\tt KE=\dfrac{1}{2}mv^2\\\\KE=\dfrac{1}{2}\times 85\times 8.5^2\\\\KE=3070.625~J$

Conservation of energy :

(KE+PE)₁ (downhill) = (KE+PE)₂ (up the hill)

PE₁=0⇒h=0

KE₂=0⇒v=0(stop), so equation becomes :

KE₁=PE₂

$\tt KE_1=3070.625\\\\KE_1=mgh_2\\\\3070.625=85\times 9.8\times h\Rightarrow h=3.686~m$

5 0

2 years ago

If 2.00 moles of H₂ and 1.55 moles of O₂ react how many moles of H₂O can be produced in the reaction below?

jekas [21]

Answer:

2 mol H₂O

Explanation:

With the reaction,

2H₂(g) + O₂(g) → 2 H₂O(g)

1.55 moles of O₂ would react completely with ( 2*1.55 ) 3.1 moles of H₂. There are not as many moles of H₂, thus H₂ is the limiting reactant.

Now we <u>calculate the moles of H₂O produced</u>, <em>starting from the moles of limiting reactant</em>:

2.00 mol H₂ * $\frac{2molH_2O}{2mol H_2}$ = 2 mol H₂O

5 0

2 years ago

The Lewis diagram for BH2 is:

Luda [366]

Answer:

Electron pair geometry- trigonal planar

There is one lone pair around the boron atom

The geometry of BH2 is bent

Explanation:

The valence shell electron pair repulsion theory offers a frame work for determining the shape of molecules based on the number of electron pairs of the valence shell of the central atom in the molecule.

In BH2-, the central atom is boron. There is a lone pair on boron. Owing to the lone pair on boron, the molecular geometry of BH2 is bent.

4 0

2 years ago