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irina [24]
3 years ago
9

PLZ HELP!!!!

Chemistry
2 answers:
Klio2033 [76]3 years ago
6 0

Answer:

relative

Explanation:

There is the relative method and absolute method. Comparing the rock to other rocks is relative.

d1i1m1o1n [39]3 years ago
5 0

Answer:

I agreeee

Explanation:

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Why is green light cool
shepuryov [24]

Because

green insinuates a feeling of moving forward and gain confidence and comfort by the color

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3 years ago
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Many enzymes are inhibited irreversibly by heavy metal ions such as Hg2+ , Cu2+ , or Ag+ , which can react with essential sulfhy
3241004551 [841]

Answer:

The minimum molecular weight of the enzyme is 29.82 g/mol

Explanation:

<u>Step 1:</u> Given data

The volume of the solution = 10 ml = 10*10^-3L

Molarity of the solution = 1.3 mg/ml

moles of AgNO3 added = 0.436 µmol = 0.436 * 10^-3 mmol

<u>Step 2:</u> Calculate the mass

Density = mass/ volume

1.3mg/mL = mass/ 10.0 mL

mass = 1.3mg/mL *10.0 mL = 13mg

<u>Step 3:</u> Calculate minimum molecular weight

Molecular weight = mass of the enzyme / number of moles

Molecular weight of the enzyme = 13mg/ 0.436 * 10^-3 mmol

Molecular weight = 29.82 g/mole

The minimum molecular weight of the enzyme is 29.82 g/mol

7 0
4 years ago
In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 2.68 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what
melomori [17]

Answer:

Y=65.7\%

Explanation:

Hello,

In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:

n_{MgI_2}^{theoretical}=2.68molMg*\frac{1molMgI_2}{1molMg} =2.68molMgI_2

Thus, we compute the percent yield as:

Y=\frac{n_{MgI_2}^{real}}{n_{MgI_2}^{theoretical}} *100\%=\frac{1.76mol}{2.68mol} *100\%\\\\Y=65.7\%

Best regards.

8 0
3 years ago
A 1 mL pipette delivers a measured volume of .98 mL (calculate the percentage error)
Natalija [7]

Answer:

2%

Explanation:

.98 is 98% of one and therefore they are missing 2%

4 0
3 years ago
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How much heat energy is required to raise the temperature of 0.360 kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of co
AVprozaik [17]
MThe  heat  energy  required  to  raise  the  temperature   of  0.36Kg   of  copper   from   22 c   to  60  c  is  calculate  using  the  following  formula

MC delta T
m(mass)=  0.360kg  in  grams  =  0.360  x1000 = 360 g
  c(specific  heat  energy)  =  0.0920  cal/g.c
delta T =  60- 23  = 37  c

heat  energy is therefore=  360g   x0.0920 cal/g.c  x 37  c=  1225.44  cal

5 0
3 years ago
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