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ch4aika [34]
3 years ago
7

To cook a steak to medium-rare, it needs to have an internal temperature of 135

Chemistry
1 answer:
Nady [450]3 years ago
7 0

t<135 because it will be less then 135°

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Help please! -A compound contains 0.013 moles of carbon, 0.039 moles of hydrogen and 0.0065 moles of oxygen. Determine the empir
stepan [7]

A) C2H6O1

To find the emperical formula, divide each mole value by the smallest

For carbon, 0.013/0.0065 = 2

For hydrogen, 0.038/0.0065= 6

For oxygen, 0.0065/0.0065= 1

Emperical formula = C2H6O1

8 0
3 years ago
Read 2 more answers
How many grams of AICI: are needed to produce 150 grams of NaCl?
Anettt [7]

Answer:

18.7g

Explanation:

5 0
2 years ago
A slurry of flakes soybeans weighing a total of 100 kg contains 75 kg of inert solids and 25 kg of solution with 10 wt% oil and
lubasha [3.4K]

Answer:

the amounts and compositions of the overflow V1 and underflow L1 leaving the stage are 75kg and 125kg respectively.

Explanation:

Let state the given parameters;

Let A= solvent (hexane)

B= solid(inert soiid)

C= solvent(oil)

F_{solution} = mass of solvent + mass of oil (i.e A+C)

<u>Feed Phase:</u>

Total feed (i.e slurry of flakes soybeans)= 100kg

B= mass of solid =75 kg

F= mass of solvent + mass of oil (i.e A+C)

 = 25kg

Mass ratio of oil to solution Y_{F} =\frac{Mass C}{Mass (A+C)}

mass of oil (C) =25 × 0.1 wt = 2.5kg

mass of hexane  in feed = 25 ×  0.9 =22.5kg + 2.5 =25kg

therefore  Y_{F} = \frac{2.5}{25}

= 0.1

mass ratio of solid to solution Y_{A}  =  \frac{Mass A}{Mass (A+C)}=[tex]\frac{75}{25}

=3

<u>Solvent Phase:</u>

C= Mass of oil= 0(kg)

A= Mass of hexane = 100kg

mass of solutions = A+C = 0+100kg

solvent= 100kg

<u>Underflow:</u>

underflow = L₁ = (unknown) ???

L₁ = E₁ + B

the value of N for the outlet and underflow is 1.5 kg

i.e N₁ = \frac{mass B}{mass(A+C)}

solution in underflow E₁ = Mass (A+C)

<u>Overflow:</u>

Overflow = V₁ = (unknown) ???

solution in overflow V₁ = Mass (A+C)

This is because, B = 0 in overflow

Solid Balance: (since the solid is inert, then is said to be same in feed & underflow).

solid in feed = solid in underflow = 75

75=  E₁ × N₁

75 =  E₁ × 1.5

E₁ = 50kg

Underflow L₁ = E₁ × B

= 50 + 75

=125kg

The Overall Balance: Feed + Solvent = underflow + overflow

100 + 100 = 125 + V₁

V₁ = 75kg

5 0
3 years ago
How many moles are in 5.5 x 10-23 molecules of H2O
masha68 [24]

Answer:

0.914moles

Explanation:

The number of moles in a substance can be got by dividing the number of atoms/molecules/particles by Avagadro's constant (6.02 × 10^23).

That is;

number of moles (n) = number of atom (nA) ÷ 6.02 × 10^23

According to this question, there are 5.5 x 10-23 molecules of H2O

n = 5.5 x 10^23 ÷ 6.02 × 10^23

n = 0.914 × 10^(23-23)

n = 0.914 × 10^0

n = 0.914 × 1

n = 0.914moles

6 0
3 years ago
How many grams of a stock solution that is 92.5 percent H2SO4 by mass would be needed to make 250 grams of a 35.0 percent by mas
IrinaVladis [17]

94.6 g.  You must use 94.6 g of 92.5 % H_2SO_4 to make 250 g of 35.0 % H_2SO_4.

We can use a version of the <em>dilution formula</em>

<em>m</em>_1<em>C</em>_1 = <em>m</em>_2<em>C</em>_2

where

<em>m</em> represents the mass and

<em>C</em> represents the percent concentrations

We can rearrange the formula to get

<em>m</em>_2= <em>m</em>_1 × (<em>C</em>_1/<em>C</em>_2)

<em>m</em>_1 = 250 g; <em>C</em>_1 = 35.0 %

<em>m</em>_2 = ?; _____<em>C</em>_2 = 92.5 %

∴ <em>m</em>_2 = 250 g × (35.0 %/92.5 %) = 94.6 g

4 0
3 years ago
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