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Sever21 [200]
3 years ago
8

For each of the following numbers, determine the number of significant figures it contains, rewrite it without using scientific

notation, and determine the number of significant figures in the result. (a) 3.050 x 10^-4, (b) 4.3200 x 10^2,(c) 8.001 x 10^-7, (d) 2.006080 x 10^5, (e) 1.503 x 10^-5, (f) 6.07510 x 10^4.
Chemistry
1 answer:
soldier1979 [14.2K]3 years ago
7 0

Answer:

Explanation:

a) .0003050                    significant figure - 4

b) 432.00                          significant figure-5

c) .0000008001                significant figure -4

d) 200608.0                       significant figure -7

e) .00001503                       significant figure-4

f) 60751.0                              significant figure- 6.

zero after decimal are all significant . zero before decimal is significant only when it is preceded by a digit.All digits are significant.

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(((NEED ANSWER QUICK!!!)))<br><br> Which is the stronger conjugate base, CN- or OCN-? Explain
Shkiper50 [21]

Answer:

The stronger conjugate base will be the weaker acid; i.e., the acid with the smaller Ka-value.

Explanation:

Given conjugate base CN⁻ => weak acid => HCN =>  Ka =4.9 x 10⁻¹⁰

Given conjugate base OCN⁻ => weak acid=> HOCN => Ka = 3.5 x 10⁻⁴

Ka(HCN) << Ka(HOCN) => CN⁻ is a much stronger conjugate base than OCN⁻

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3 years ago
How many particles are in 47.7 g of Magnesium? (Round the average
Lynna [10]

Answer:

1.18 × 10²⁴ particles Mg

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

<u>Step 1: Define</u>

47.7 g Mg

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of Mg - 24.31 g/mol

<u>Step 3: Convert</u>

<u />47.7 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg} )(\frac{6.022 \cdot 10^{23} \ particles \ Mg}{1 \ mol \ Mg} ) = 1.18161 × 10²⁴ particles Mg

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

1.18161 × 10²⁴ particles Mg ≈ 1.18 × 10²⁴ particles Mg

3 0
2 years ago
Estimate the Calorie content of 65 g of candy from the following measurements. A 15-g sample of the candy is placed in a small a
GrogVix [38]

Answer:

The calorie content of  65g of candy is 326.78 cal

Explanation:

Step 1: Data given

Mass of the candy = 15.00 grams

Mass of the container = 0.325 kg

Mass of water = 1.75kg

0.624 kg at an initial temperature of 15.0°C.

The specific heat of aluminium = 0.22 Cal/kg°C

The specific heat of water = 1 cal/kg°C

Step 2: Calculate calorie content for a 15 gram sample

ΔQ = Σm*c*ΔT

 ⇒ m = mass in grams

⇒ with c= the specific heat in Cal/kg°C

⇒ with ΔT = T2 -T1 = the change in temperatures in °C

ΔQ = m(bomb) * C(aluminium) * ΔT + m(cup) * C(aluminium) * ΔT + m(H2O) * c(H20) * ΔT

ΔQ = (m(bomb) + m(cup)) * c(aluminium)  + m(H2O)*c(H20) ) * ΔT

⇒ with mass of the bomb calorimeter = 0.325 kg

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⇒ with c(aluminium) = the specific heat of aluminium = 0.22 Cal/kg°C

⇒ with mass of water = 1.75 kg

⇒ with c(water) = the heat capacity of water = 1 Cal/kg°C

⇒ with ΔT = the change in temperature = T2 - T1 = 53.5 - 15.0 = 38.5 °C

ΔQ = 0.325*0.22*38.5 + 0.624*0.22*38.5 + 1.75*1*38.5

ΔQ = ((0.325 + 0.624)*0.22 + 1.75*1)*38.5

ΔQ = 75.41 cal

Step 3: Calculate the calorie content for a 65 gram sample

For a 65g sample the calorie content will be more or less 4x higher than a 15 gram sample:

ΔQ = 75.41  * (65/15) = 326.78 cal

8 0
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