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4 years ago

Which of the following prefixes would be best to use when measuring the distance from your home to the store?

1 answer:

4 0

I would say the most correct answer is Kilo, as in kilometers. The store would certainly not be milli, or millimeters away from you, would it now? :)

Hope this helps!

You might be interested in

(ASAP) 25 cm cubed of the champagne were placed in a conical flask. Describe how the volume of sodium hydroxide solution needed

Zielflug [23.3K]

Answer:

in a conical flask. Describe how the volume of sodium hydroxide solution needed to react completely with the weak acids in 25cm cubed of this champagne can be found by titration, using phenolphthalein indicator.

5 0

3 years ago

One of the nuclides in each of the following pairs is radioactive. Predict which is radioactive and which is stable.a. 39/19K an

marishachu [46]

Answer:

a) 39/19 K : stable nuclide, 40/19 K : radioactive nuclide.

b) 209B: stable nuclide, 208Bi : radioactive nuclide

c) nickel-58 : stable nuclide, nickel-65 : radioactive nuclide.

Explanation:

As per the rule, nuclides having odd number of neutrons are generally not stable and therefore, are radioactive.

Mass number (A) = Atomic number (Z) + No. of neutrons (N)

Or, N = A - Z

39/19 K and 40/19 K

Calculate no. of neutrons in 39/19 K as follows:

atomic no. = 19, mass no. 39

N = 39 - 19

= 20 (even no.)

Calculate no. of neutrons in 40/19 K as follows:

atomic no. = 19, mass no. 40

N = 40 - 19

= 21 (odd no.)

Therefore, 39/19 K is a stable nuclide and 40/19 K is a radioactive

nuclide.

209Bi and 208Bi

Calculate no. of neutrons in 209Bi as follows:

atomic no. = 83, mass no. 209

N = 209 - 83

= 126 (even no.)

Calculate no. of neutrons in 208Bi as follows:

atomic no. = 83, mass no. 208

N = 208 - 83

= 125 (odd no.)

Therefore, 209Bi is a stable nuclide and 208Bi is a radioactive nuclide.

nickel-58 and nickel-65

Calculate no. of neutrons in nickel-58 as follows:

atomic no. = 28, mass no. 58

N = 58 - 28

= 30 (even no.)

Calculate no. of neutrons in nickel as follows:

atomic no. = 28, mass no. 65

N = 65 - 28

= 37 (odd no.)

Therefore,nickel-58 is a stable nuclide and nickel-65 is a radioactive nuclide.

5 0

3 years ago

In the following reaction, how many grams of nitroglycerin C3H5(NO3)3 will decompose to give 25 grams of CO2?

gogolik [260]

4C₃H₅(NO₃)₃ $_{(l)}$ ------> 12CO₂ $_{(g)}$ + 6N₂ $_{(g)}$ + 10H₂O $_{(g)}$ + O₂ $_{(g)}$

mol of CO₂ = $\frac{mass}{molar mass}$

= $\frac{25g}{44.01 g/mol}$

mol ratio of CO₂ : C₃H₅(NO₃)₃

12 : 4

∴ if mole of CO₂ = 0.568 mol

then " " C₃H₅(NO₃)₃ = $\frac{0.568 mol}{12} * 4$

= 0.189 mol

∴ mass of nitroglycerin = mole * Mr

= 0.189 mol * 227.0995 g / mol

= 43.00 g

8 0

3 years ago

A 100g of sample of a compound is combusted in excess oxygen and the products are 2.492g of CO2 and 0.6495 of H2O. Determine the

ruslelena [56]

The empirical formula : C₁₁O₁₄O₃

<h3>Further explanation</h3>

The assumption of the compound consists of C, H, and O

mass of C in CO₂ =

$\tt \dfrac{12}{44}\times 2.492=0.680~g$

mass of H in H₂O =

$\tt \dfrac{2.1}{18}\times 0.6495=0.072~g$

mass of O :

mass sample-(mass C + mass H)

$\tt 1-(0.68+0.072)=0.248`g$

mol of C :

$\tt \dfrac{0.68}{12}=0.056$

mol of H :

$\tt \dfrac{0.072}{1}=0.072$

mol of O :

$\tt \dfrac{0.248}{16}=0.0155$

divide by 0.0155(the lowest ratio)

C : H : O ⇒

$\tt \dfrac{0.056}{0.0155}\div \dfrac{0.072}{0.0155}\div \dfrac{0.0155}{0.0155}=3.6\div 4.6\div 1\\\\\dfrac{11}{3}\div \dfrac{14}{3}\div \dfrac{3}{3}=11:14:3$

3 0

3 years ago

BRAINLIESTT ASAP! PLEASE HELP :)

Nadya [2.5K]

Hello, and this is my own words Periodic trends are specific patterns in the properties of chemical elements that are revealed in the periodic table of elements. Major periodic trends include electronegativity, ionization energy, electron affinity, atomic radii, ionic radius, metallic character, and chemical reactivity.

Hope This Helps. and have good day!

3 0

3 years ago