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3 years ago

0.0006510 kg has how many significant figures.

Chemistry

2 answers:

prisoha [69]3 years ago

6 0

Answer:

4 significant figures as zeros after decimal should not be counted

Explanation:

Please mark brainliest and have a great day!

Nady [450]3 years ago

6 0

Answer:

4 significant figures

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All of the following are examples of commonly used tools in relief printing, except which?

guapka [62]

All are used in relief printing except a gouge "D". A gouge is a concave bladed chisel used in sculpting. Lead can be the medium that the relief is in. Paper is the material that the final print is on. A mat knife is the blade used to cut the paper to the appropriate size.

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3 years ago

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A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-

Komok [63]

Answer: The cell potential of the cell is +0.118 V

Explanation:

The half reactions for the cell is:

Oxidation half reaction (anode): $Ni(s)\rightarrow Ni^{2+}+2e^-$

Reduction half reaction (cathode): $Ni^{2+}+2e^-\rightarrow Ni(s)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = ?

$[Ni^{2+}_{diluted}]$ = $1.00\times 10^{-4}M$

$[Ni^{2+}_{concentrated}]$ = 1.0 M

Putting values in above equation, we get:

$E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}$

$E_{cell}=0.118V$

Hence, the cell potential of the cell is +0.118 V

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3 years ago

A liquid dietary supplement is packaged in 10-ml dropper containers to deliver 2000 international units of vitamin d3 in each dr

RoseWind [281]

Answer : 37 drops are delivered per milliliter of the solution.

Explanation :

The problem gives us lot of extra information.

We want to find the number of drops delivered in 1 milliliter here.

We have been given that, one drop of the solution delivers 0.027 mL of solution.

Let us use this as a conversion factor, $\frac{1 drop}{0.027 mL}$

Let us find number of drops in 1 mL using this conversion factor.

$1 mL \times \frac{1 drop}{0.027mL} = 37.0 drops$

Therefore we can say that 37 drops are delivered per milliliter of the solution.

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4 years ago

What happens to the reaction rate as a reactant gets used up?

Oxana [17]

Answer: B. The rate goes down

Explanation:

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3 years ago

A chemist must prepare 800.0mL of sodium hydroxide solution with a pH of 12.10 at 25°C. She will do this in three steps: Fill a

AnnZ [28]

Answer:

0.42 g

Explanation:

We have:

pH = 12.10 (25 °C)

V = 800.0 mL = 0.800 L

To find the mass of sodium hydroxide (NaOH) we can use the pH:

$14 = pH + pOH$

$pOH = 14 - pH = 14 - 12.10 = 1.90$

$pOH = -log ([OH^{-}])$

$[OH]^{-} = 10^{-pOH} = 10^{-1.90} = 0.013 M$

Now, we can find the number of moles (η) of OH:

$\eta = ([OH]^{-})*V = 0.013 mol/L * 0.800 L = 1.04 \cdot 10^{-2} moles$

Since we have 1 mol of OH in 1 mol of NaOH, the number of moles of NaOH is equal to 1.04x10⁻² moles.

Finally, with the number of moles we can find the mass of NaOH:

$m = \eta * M$

Where M is the molar mass of NaOH = 39.9 g/mol

$m = 1.04 \cdot 10^{-2} moles * 39.9 g/mol = 0.42 g$

Therefore, the mass of sodium hydroxide that the chemist must weigh out in the second step is 0.42 g.

I hope it helps you!

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3 years ago