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KatRina [158]
2 years ago
10

PLEASE HELP

Chemistry
1 answer:
devlian [24]2 years ago
6 0
Metals have a low electron affinity- a less likely chance to gain electrons because they want to give up their valence electrons rather than gain electrons, which require more energy than necessary.
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Given the balanced equation, calculate the mass of product that can be prepared from 2.36 g of zinc metal.
Nataly_w [17]
Moles of Zn present= 2.36/65.4= 0.0361 moles
Therefore maximum moles of ZnO= 0.0722
Mass of one mole of ZnO= 81.4
Mass of ZnO produced= 0.0722 x 81.4= 5.87g
6 0
3 years ago
I need to check some chemistry questions. Help with any question is appreciated! :) Please include an explanation with your conc
finlep [7]

1. 4.67 kg; 2. 4.8 ×10^5 kg; 3. 0.106 cm^3; 4. 1.7 g/cm^3

<em>Q1. Mass of Hg </em>

Mass = 345 mL × (13.53 g/1 mL) = 4670 g = 4.67 kg

<em>Q2. Mass of Pb </em>

<em>Step 1</em>. Calculate the <em>volume of the Pb</em>.

<em>V = lwh</em> = 6.0 m × 3.5 m × 2.0 m = 42.0 m^3

<em>Step 2</em>. Calculate the <em>mass of the Pb</em>.

Mass = 42.0 m^3 × (11 340 kg/1 m^3) = 4.8 × 10^5 kg

<em>Q3. Volume of displaced water </em>

Volume of Ag = 0.987 g × (1 cm^3/9.320 g) = 0.106 cm^3

<em>Archimedes</em>: volume of displaced water = volume of Ag = <em>0.106 cm^3</em>

<em>4. Density of metal </em>

<em>Step 1</em>. Convert <em>ounces to grams </em>

Mass = 3.35 oz × (28.35 g/1 oz) = 94.97 g

<em>Step 2</em>. Calculate the <em>volume in cubic inches </em>

<em>V = lwh</em> = 3.0 in × 2.5 in × 0.45 in = 3.38 in^3

<em>Step 3</em>. Convert <em>cubic inches to cubic centimetres</em><em> </em>

<em>V</em> = 3.38 in^3 × (2.54 cm/1 in)^3 = 55.3 cm^3

<em>Step 4</em>. Calculate the <em>density</em>

ρ = <em>m</em>/<em>V</em> = (94.97 g/55.3 cm^3) = 1.7 g/cm^3 (magnesium?)

4 0
3 years ago
El numero de moles en 70 gramos de HF
saveliy_v [14]

Answer:

enviornmental health perspectives

Explanation:

7 0
3 years ago
Read 2 more answers
11. What is the maximum number of electrons that an atomic orbital can contain?
Arada [10]

Answer:

2

Explanation:

Each orbital can hold two electrons. One spin-up and one spin-down.

6 0
3 years ago
A compound of a transition metal and iodine is 56.7% metal by mass.How many grams of the metal can be obtained from 630 g of thi
erma4kov [3.2K]

Answer:

357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine

Explanation:

In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.

Given mass of sample compound = 630 g

Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;

56.7 % = 56.7/100 = 0.567

Mass of transition metal = 0.567 * 630 = 357.21 g

Therefore, the mass of the transition metal  present in 630 g of the compound is approximately 357 g

4 0
3 years ago
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