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Serga [27]
3 years ago
5

Anomalous data shows on a graph as a

Chemistry
1 answer:
KiRa [710]3 years ago
6 0
Anomalous data on a graph would show up as say a very high or very low value which does not fit in with the normal values which may be background values.If it was a straight line graph then the anomalous point would plot well above or below the line or if it was a bar graph ie a histogram it would be much higher or lower than the surrounding data. In mineral exploration, anomalies are looked for in say geophysics or geochemistry data values for high or low magnetism or conductivity or high chemical values indicating the presence of valuable minerals at that point.
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Determining Density and Using Density to Determine Volume or Mass
Shalnov [3]

corrected question:

Determining Density and Using Density to Determine Volume or Mass

(a) Calculate the density of mercury if 1.00 × 10 g occupies a volume of 7.36 cm³

(b) Calculate the volume of 65.0 g of liquid methanol (wood alcohol) if its density is 0.791 g/mL.

(c) What is the mass in grams of a cube of gold (density = 19.32 g/cm) if the length of the cube is 2.00 cm?

(d) Calculate the density of a 374.5-g sample of copper if it has a volume of 41.8 cm³ A student needs 15.0 g of ethanol for an experiment. If the density of ethanol is 0.789 g/mL, how many milliliters of ethanol are needed? What is the mass, in grams, of 25.0 mL of mercury (density = 13.6 g/mL)?

Answer:

density = \frac{mass}{volume}

ρ=m/v ,m=ρv,    v=m/ρ

(a)m=1*10g  , v=7.36cm³

    ρ=10/7.36 =1.36g/cm³

(b) m=65g, ρ=0.791 g/mL.

   v= 65/0.791 =82.17g/mL

(c) ρ=19.32g/cm³, l=2cm, v=l³=8cm³

    m=19..32*8=154.56g/cm³

(d) mass of copper=374.5g , v=41.8cm³

   ρ=374.5/41.8 =8.96g/cm³

 mass of ethanol=15g,  density of ethanol=0.789g/mL

v=15/0.789 =19.01mL

volume of mecury=25mL, density of mercury=13.6g/mL

m=25*13.6=340g

4 0
3 years ago
Can someone please help me
lilavasa [31]

Answer:

Z was smaller than X

Explanation:

if X<Y and Y<Z

therefore X<Z

8 0
3 years ago
The density of an object is calculated as Density = . If 100mL of water weighs 100g calculate the density?
iren2701 [21]
D = mass / volume

d = 100 g / 100 mL

d = 1.0 g/mL
3 0
3 years ago
Which of the following equations are correctly balanced?
AlladinOne [14]

i think it's I

I was confused by IV then search on gg and it said ZnSO4 should be Zn2SO4 instead but still im not sure Zn2SO4 is real

3 0
3 years ago
A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an
erma4kov [3.2K]

Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

7 0
3 years ago
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