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3 years ago

14

Which of the measurement contain three significant figure?

1 answer:

Hunter-Best [27]3 years ago

4 0

Answer:

The answer is explained below

Explanation:

The question is not complete.

The correct question should be like:

Which of the measurement contain three significant figure?

500

395

0.001

0.00100

1 × 10⁻³

5 × 10²

Answer:

The following are rules applied when considering significant figures:

i) All digits are significant excluding zero. E.g of significant digits are 1,2,3,4,5,6,7,8,9

ii) Any zero that appear within or after non zero digits are significant e.g. 200, 1004, 500009

iii) Zeros before non zero digits are not significant. An example shows significant digits bolded 001, 0.0012, 001.56

iv) Zeros that come after non zero digits are significant even if the number is in decimal form. An example shows significant digits bolded 1000, 0.001200, 001.5600

Therefore:

500: It contains three significant digits

395: It contains three significant digits

0.001: It contains only one significant digits

0.00100: It contains three significant digits

1 × 10⁻³: 1 × 10⁻³ = 0.001. It contains only one significant digits

5 × 10²: 5 × 10² = 500. It contains three significant digits

The measurements that contain three significant figure are 500, 395, 0.00100 and 5 × 10²

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Give the half equation to show what happens to oxygen when it is rusting

ycow [4]

Answer:

½O 2 + 2e - + H 2O → 2OH.

Explanation:

Redox reactions - Higher

In terms of electrons:

oxidation is loss of electrons

reduction is gain of electrons

Rusting is a complex process. The example below show why both water and oxygen are needed for rusting to occur. They are interesting examples of oxidation, reduction and the use of half equations:

iron loses electrons and is oxidised to iron(II) ions: Fe → Fe2+ + 2e-

oxygen gains electrons in the presence of water and is reduced: ½O2 + 2e- + H2O → 2OH-

iron(II) ions lose electrons and are oxidised to iron(III) ions by oxygen: 2Fe2+ + ½O2 → 2Fe3+ + O2-

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3 years ago

According to solubility rules, which of the following compounds is soluble in water?

Nata [24]

The answer is A hope this help

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3 years ago

Calculate the change in entropy when 1.00 kg of water at 100 ∘C is vaporized and converted to steam at 100 ∘C. Assume that the h

andrew11 [14]

Answer : The change in entropy is $6.05\times 10^3J/K$

Explanation :

Formula used :

$\Delta S=\frac{m\times L_v}{T}$

where,

$\Delta S$ = change in entropy = ?

m = mass of water = 1.00 kg

$L_v$ = heat of vaporization of water = $2256\times 10^3J/kg$

T = temperature = $100^oC=273+100=373K$

Now put all the given values in the above formula, we get:

$\Delta S=\frac{(1.00kg)\times (2256\times 10^3J/kg)}{373K}$

$\Delta S=6048.25J/K=6.05\times 10^3J/K$

Therefore, the change in entropy is $6.05\times 10^3J/K$

5 0

3 years ago

Calculate the radius ratio for NaBr if the ionic radii of Na + and Br − are 102 pm and 196 pm , respectively. radius ratio: Base

Fudgin [204]

Answer : The expected coordination number of NaBr is, 6.

Explanation :

Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.

This is represented by,

$\frac{r_{cation}}{r_{anion}}$

When the radius ratio is greater than 0.155, then the compound will be stable.

Now we have to determine the radius ration for NaBr.

Given:

Radius of cation, $Na^+$ = 102 pm

Radius of cation, $Br^-$ = 196 pm

$\frac{r_{cation}}{r_{anion}}=\frac{102}{196}=0.520$

As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.

The relation between radius ratio and coordination number are shown below.

Therefore, the expected coordination number of NaBr is, 6.

8 0

3 years ago

Leo carefully pipets 50.0 mL of 0.500 M NaOH into a test tube. She places the test tube

Bess [88]

Let's consider the neutralization reaction between HCl and NaOH.

NaOH + HCl ⇒ NaCl + H₂O

To determine the pH of the resulting mixture, we need to determine the reactant in excess. First, we will calculate the reacting moles of each reactant.

NaOH: 0.0500 L × 0.500 mol/L = 0.0200 mol

HCl: 0.0750 L × 0.250 mol/L = 0.0188 mol

Now, let's determine the reactant in excess and the remaining moles of that reactant.

NaOH + HCl ⇒ NaCl + H₂O

Initial 0.0200 0.0188

Reaction -0.0188 -0.0188

Final 1.20 × 10⁻³ 0

The volume of the mixture is 50.0 mL + 75.0 mL = 125.0 mL. Then, 1.20 × 10⁻³ moles of NaOH are in 125.0 mL of solution. The concentration of NaOH is:

[NaOH] = 1.20 × 10⁻³ mol/0.1250 L = 9.60 × 10⁻³ M

NaOH is a strong base according to the following equation.

NaOH ⇒ Na⁺ + OH⁻

The concentration of OH⁻ is 1/1 × 9.60 × 10⁻³ M = 9.60 × 10⁻³ M.

The pOH is:

pOH = -log [OH⁻] = -log 9.60 × 10⁻³ = 2.02

We will calculate the pH using the following expression.

pH = 14.00 - pOH = 14.00 - 2.02 = 11.98

The pH is 11.98. Since pH > 7, the solution is basic.

You can learn more about neutralization here: brainly.com/question/16255996

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3 years ago