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Anika [276]
3 years ago
8

I need help with part B . please

Chemistry
1 answer:
myrzilka [38]3 years ago
8 0
So it's good to map out what you know you have and work from there:
We have two liter measurements and one mole measurement, and we need to find the moles.

For this problem, think of it this way: 46 liters of gas = 1.4 moles.
If one side changes, the other has to as well (if the liters decrease, the moles decrease. if the liters increase, so do the moles.) What you can do is put this into a fraction:

  <span><u>1.4 moles</u></span>
      46 L   <span> </span>

if we know that each liter of gas is equal to x amount of moles, we know that 11.5 liters equals some amount of moles. You can put this into a fraction too, and make it equal to the other fraction:

   <span><u>1.4 moles</u></span> = <u>x moles</u> 
         46 L         11.5 L

Then get your calculator out and do some algebra.

11.5 * (1.4/46) = x

The answer should come out to be: 0.35 moles
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What orbital do the transition metals finish their electron configuration in
PIT_PIT [208]

Answer:

d orbitals

Explanation:

Transition  metals are generally known as d-block elements. The electronic configuration of all transition elements finish in a d-orbital weather they are first row, second row or third row transition elements. This is the thread that holds all the elements of the transition series together.

This is why elements of the transition series are generally called the d-block elements.

5 0
3 years ago
A mixture of methanol and methyl acetate contains 15.0 wt% methanol.
Usimov [2.4K]

The mixture flow rate in lbm/h = 117.65 lbm/h

<h3>Further explanation</h3>

Given

15.0 wt% methanol

The flow rate of the methyl acetate :100 lbm/h

Required

the mixture flow rate in lbm/h

Solution

mass of methanol(CH₃OH, Mw= 32 kg/kmol) in mixture :

\tt 15\%\times 200~kg=30~kg\\\\mol=\dfrac{mass}{MW}=\dfrac{30~kg}{32~kg/kmol}=0.9375~kmol

mass of the methyl acetate(C₃H₆O₂,MW=74 kg/kmol,85% wt) in 200 kg :

\tt 85\%\times 200=170~kg\\\\mol=\dfrac{170}{74}=2.297~kmol

Flow rate of the methyl acetate in the mixture is to be 100 lbm/h.

1 kg mixture = 0.85 .methyl acetate

So flow rate for mixture :

\tt \dfrac{1~kg~mixture}{0.85~methyl~acetat}\times 100~lbm/h=117.65~lbm/h

5 0
2 years ago
If an object has a density of 0.55 g/mL, what is its density in cg/L?
expeople1 [14]

"cg" is centigram, which is one-hundredth of a gram.

I will first convert from g to cg (multiply by 100), then from mL to L (multiply by 1000).

\frac{0.55g}{mL}*\frac{100cg}{1g}*\frac{1000mL}{1L}=55,000\frac{cg}{L} \ or \ 5.5e4\frac{cg}{L}

3 0
3 years ago
if the lightbulb receives 100 J of electrical energy, and gives off 75 energy, how much heat (thermal energy away from the light
oksano4ka [1.4K]

Answer:

Amount of heat energy released by light bulb = 25 joules

Explanation:

Given:

Energy receive by light bulb = 100 Joules

Energy released by light bulb as light energy = 75 Joules

Find:

Amount of heat energy released by light bulb

Computation:

We know that, energy is neither be created nor destroys

So,

Using Law of conservation of energy

Energy receive by light bulb = Energy released by light bulb as light energy + Amount of heat energy released by light bulb

100 = 75 + Amount of heat energy released by light bulb

Amount of heat energy released by light bulb = 100 - 75

Amount of heat energy released by light bulb = 25 joules

8 0
3 years ago
CaC12 * 3H20 is correctly named
ki77a [65]

Answer:

calcium chloride deihydrate

4 0
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