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3 years ago

How do I identify significant figures in chemistry?

1 answer:

3 0

1) START counting for sig. figs. On the FIRST non-zero digit.

2) STOP counting for sig. figs. On the LAST non-zero digit.

3) Non-zero digits are ALWAYS significant.

4) Zeroes in between two non-zero digits are significant. All other zeroes are insignificant.

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A student weighed out a 2.055 g sample of a cobalt chloride hydrate, ConClmpH2O, where n, m, and p are integer values to be dete

topjm [15]

The given mass of cobalt chloride hydrate = 2.055 g

A sample of cobalt chloride hydrate was heated to drive off waters of hydration and the anhydrate was weighed.

The mass of anhydrous cobalt chloride = 1.121 g anhydrate.

The mass of water lost during heating = 2.055 g - 1.121 g = 0.934 g

Converting mass of water of hydration present in the hydrate to moles using molar mass:

Mass of water = 0.934 g

Molar mass of water = 18.0 g/mol

Moles of water = $0.934 g * \frac{1 molH_{2}O }{18 g H_{2}O } =0.0519 mol H_{2}O$

8 0

3 years ago

Pls help with science

MissTica

The answer is C because it is believed that those two were once connected

6 0

3 years ago

Change 0.00765 kL into mL

bagirrra123 [75]

Answer:

7650

Explanation:

formula- multiply the volume value by 1e+6

4 0

4 years ago

Read 2 more answers

Use the Gizmo to estimate the age of each of the objects below. For these questions, each second in the Gizmo represents 1,000 y

ziro4ka [17]

Answer:

Check the explanation

Explanation:

AT = A0 e(-T/H)

... where A0 is the starting activity, AT is the activity at some time T, and H is the half-life, in units of T.

Substituting what we know, we get...

0.71 = (1) e(-T/5730)

Solve for T...

loge(0.71) = -T/5730

T = -loge(0.71)(5730)

T = 1962 (conservatively rounded, T = 2000)

similarly for all

for aboriginal charcoal

0.28 = (1) e(-T/5730)

Solve for T...

loge(0.28) = -T/5730

T = -loge(0.28)(5730)

T = 7294 (conservatively rounded, T = 7000)

for mayan headdress

0.89 = (1) e(-T/5730)

Solve for T...

loge(0.89) = -T/5730

T = -loge(0.89)(5730)

T = 667 (conservatively rounded, T = 700)

for neanderthal

0.05 = (1) e(-T/5730)

Solve for T...

loge(0.05) = -T/5730

T = -loge(0.05)(5730)

T = 17165 (conservatively rounded, T = 17000)

7 0

3 years ago

An electrochemical cell has the following standard cell notation.

icang [17]

The cell notation is:

$Mg(s)|Mg^{+2}(aq)||Ag^{+}(aq)|Ag(s)$

here in cell notation the left side represent the anodic half cell where right side represents the cathodic half cell

in anodic half cell : oxidation takes place [loss of electrons]

in cathodic half cell: reduction takes place [gain of electrons]

1) this is a galvanic cell

2) the standard potential of cell will be obtained by subtracting the standard reduction potential of anode from cathode

$E^{0}_{Mg}=-2.38V$

$E^{0}_{Ag}=+0.80V$

Therefore

$E^{0}_{cell}=0.80-(-2.38)=+3.18V$

3) as the value of emf is positive the reaction will be spontaneous as the free energy change of reaction will be negative

Δ $G^{0}=-nFE^{0}$

As reaction is spontaneous and there will be conversion of chemical energy to electrical energy it is a galvanic cell.

7 0

3 years ago