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erastovalidia [21]
3 years ago
11

What element is in the same family as Pb but it has fewer protons than Na​

Chemistry
2 answers:
Fantom [35]3 years ago
8 0

Answer:

Carbon

Explanation:

Shtirlitz [24]3 years ago
4 0
Carbon!!!!!!!!!!!!!!!!!!
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How do you define an anion?
Elis [28]

An anion has more electrons than protons, it is a negatively charged ion.

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3 years ago
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What is the moisture content of hard gelatin capsule?​
insens350 [35]

Answer:

The standard moisture content specification for hard gelatin capsules is between 13 % w/w and 16 % w/w.  This value can vary depending upon the conditions to which they are exposed: at low humidity's they will lose moisture and become brittle, and at high humidity's they will gain moisture and soften.

Explanation:

Hope this helps!

6 0
3 years ago
How many moles of ammonia are there in a 346 g sample of pure nh3?
Citrus2011 [14]
Molar mass NH3 = 17.031 g/mol

1 mole NH3 -------------- 17.031 g
?? moles NH3 ---------- 346 g

346 x 1 / 17.031

=> 20.31 moles of NH3

6 0
3 years ago
What is the minimum mass of magnetite (Fe,O2), a iron ore, from which 120. g of pure iron could be extracted? Be sure your answe
Schach [20]

Answer:

166 g

Explanation:

Step 1: Write the reaction for the obtaining of Fe from magnetite

Fe₃O₄ ⇒ 3 Fe + 2 O₂

Step 2: Calculate the moles corresponding to 120 g of Fe

The molar mass of Fe is 55.85 g/mol.

120 g × (1 mol/55.85 g) = 2.15 mol

Step 3: Calculate the moles of Fe₃O₄ required to produce 2.15 moles of Fe

The molar ratio of Fe₃O₄ to Fe is 1:3. The moles of Fe₃O₄ required are 1/3 × 2.15 mol = 0.717 mol

Step 4: Calculate the mass corresponding to 0.717 moles of Fe₃O₄

The molar mass of Fe₃O₄ is 231.53 g/mol.

0.717 mol × 231.53 g/mol = 166 g

6 0
3 years ago
Prior to hosting an international soccer match, the local soccer club needs to replace the artifical turf on their field with gr
GuDViN [60]

Answer:

It will cost $68,620.5 to the club to add the grass turf to their field.

Explanation:

length of the field = l = 0.102 km = 0.102 × 1000 m

( 1km = 1000 m)

Width of the field w = 0.069 km = 0.069 × 1000 m

Area of the field , A= l × w

A=0.102\times 1000 m\times 0.069\times 1000 m=7,038 m^2

Cost of grass turfing = 9.75 \$/m^2

Cost of grass turfing on field of 7,038 m^2 :

=7,038 m^2\times 9.75 \$/m^2=\$68,620.5

It will cost $68,620.5 to the club to add the grass turf to their field.

4 0
3 years ago
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