All categories

3 years ago

How many significant figures are in each measurement? 0.00052 m

Chemistry

2 answers:

NikAS [45]3 years ago

8 0

All the measurement contain two significant figures

leonid [27]3 years ago

6 0

a. 0.00052 m

Answer: two significant figures.

Explanation:

The zeros to the left of 52 do not count as signficant figures.

Only, 5 and 2 are significant figures.

b. 9.8 x 10⁴ g

Answer: two significant figures.

Explanation:

Both 9 and 8 count as significant figures. Here there are not additional zeros that might confuse you.

c. 5.050 mg

Answer: four significant figures.

Explanation:

The zeros to the right count as signficiant figjres. So here, 5, 0, 5, and 0 are the significant figures.

d. 8700 ml

Answer: two signficant figures.

Explanation:

Since the two zeros to the right are used to determine the place value of 87, you cannot count them as significant figures. Here, only 8 and 7 are significant figures.

You might be interested in

Cobalt-60 is a strong gamma emitter that has a half- life of 5.26 yr. The co balt-60 in a radiotherapy unit must be replaced whe

Alenkasestr [34]

<u>Answer:</u> The sample of Cobalt-60 isotope must be replaced in January 2027

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life of the reaction = 5.26 years

Putting values in above equation, we get:

$k=\frac{0.693}{5.26yrs}=0.132yrs^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $0.132yr^{-1}$

t = time taken for decay process = ? yr

$[A_o]$ = initial amount of the sample = 100 grams

[A] = amount left after decay process = (100 - 75) = 25 grams

Putting values in above equation, we get:

$0.132=\frac{2.303}{t}\log\frac{100}{25}\\\\t=10.5yrs$

The original sample was purchased in June 2016

As, June is the 6th month of the year, which means the time period will be $2016+\frac{6}{12}=2016.5$

Adding the time in the original time period, we get:

$2016.5+10.5=2027$

Hence, the sample of Cobalt-60 isotope must be replaced in January 2027

3 0

3 years ago

Predict the products of each of these reactions and write balanced complete ionic and net ionic equations for each. If no reacti

Bumek [7]

Answer:

Explanation:

Part A : LiCl(aq) + AgNO₃(aq)→

Chemical equation:

LiCl(aq) + AgNO₃(aq) → AgCl(s) + LiNO₃(aq)

Ionic equation:

Li⁺(aq) + Cl⁻(aq) + Ag⁺(aq) + NO₃⁻(aq) → AgCl(s) + Li⁻(aq) + NO⁻₃(aq)

Net ionic equation:

Cl⁻(aq) + Ag⁺(aq) → AgCl(s)

C = H2SO4(aq)+Li2SO3(aq)→

Chemical equation:

H₂SO₄(aq) + Li₂SO₃(aq) → Li₂SO₄(aq) + SO₂(g) + H₂O(l)

Ionic equation:

2H⁺(aq) + SO²⁻₄(aq) + 2Li⁺(aq) + SO₃²⁻(aq) → 2Li⁺ (aq) + SO₄²⁻(aq) + SO₂(g) + H₂O(l)

Net ionic equation:

2H⁺ + SO₃²⁻(aq) → SO₂(g) + H₂O(l)

Part E: HClO4(aq)+Ca(OH)2(aq)→

Chemical equation:

HClO₄(aq) + Ca(OH)₂(aq) → Ca(ClO₄)₂ (aq) + H₂O(l)

Balanced Chemical equation:

2HClO₄(aq) + Ca(OH)₂(aq) → Ca(ClO₄)₂ (aq) + 2H₂O(l)

Ionic equation:

2H⁺(aq) + 2ClO⁻₄(aq) + Ca²⁺(aq) + (OH)²⁻₂(aq) → Ca²⁺(aq) +(ClO₄)²⁻₂ (aq) + 2H₂O(l)

Net ionic equation:

2H⁺(aq) + (OH)²⁻₂(aq) → 2H₂O(l)

Part F: Cr(NO3)3(aq)+LiOH(aq)→

Chemical equation:

Cr(NO₃)₃(aq) + LiOH (aq) → LiNO₃(aq) + Cr(OH)₃(s)

Balanced chemical equation;

Cr(NO₃)₃(aq) + 3LiOH (aq) → 3LiNO₃(aq) + Cr(OH)₃(s)

Ionic equation:

Cr³⁺(aq) + 3NO₃⁻(aq) + 3Li⁺(aq) + 3OH⁻ (aq) → 3Li⁺(aq) + 3NO⁻₃(aq) + Cr(OH)₃(s)

Net ionic equation:

Cr³⁺(aq) + 3OH⁻ (aq) → Cr(OH)₃(s)

Part H: HCl(aq)+Hg2(NO3)2(aq)→

Chemical equation:

HCl (aq) + Hg₂(NO₃)₂(aq) → Hg₂Cl₂ (s) + HNO₃(aq)

Balanced chemical equation:

2HCl (aq) + Hg₂(NO₃)₂(aq) → Hg₂Cl₂ (s) + 2HNO₃(aq)

Ionic equation;

2H⁺(aq) + 2Cl⁻ (aq) + 2Hg⁺(aq) + 2NO₃⁻(aq) → Hg₂Cl₂ (s) + 2H⁺(aq) + 2NO⁻₃(aq)

Net ionic equation:

2Cl⁻ (aq) + 2Hg⁺(aq) → Hg₂Cl₂ (s)

8 0

3 years ago

Phenolphthalein indicator was added, and the solution in the flask was titrated with 0.215M NaOH until the indicator just turned

NNADVOKAT [17]

In titration, the moles of acid equal moles of base. You were given that 22.75ml of 0.215M NaOH is used, so calculate the number of moles of that base the experiment used in total. After that because you know mol base = mol acid, whatever amount of base you use must be the total amount of acid present in the solution. You were given the volume of the acid, and you have just found the total mols of acid. Using these two information, solve for the concentration. And one more thing, even though I'm pretty sure it won't affect your answer, you should always convert things to the proper units. Since the concentration we're talking about in this problem is molarity, which has the unit mol/L, you should always have all of your numbers in these units. It just make it simpler and will not confuse you

3 0

3 years ago

Read 2 more answers

What are crayons made of?

Mashutka [201]

They are made of different color of waxes

7 0

3 years ago

Read 2 more answers

What are the different acid bases

Alex

Answer: