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lbvjy [14]
3 years ago
12

Which substance tends to form anions when bonding with other elements?

Chemistry
2 answers:
weqwewe [10]3 years ago
7 0

Answer:

B) Oxygen

Explanation:

i just took the test.

alekssr [168]3 years ago
5 0

Oxygen or podium or calcium

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Why does the same side of the moon always face earth? A. The rotational period of earth is the same as that of the moon B. The r
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Answer: A. the rotational period of the earth is the same as that of the moon

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3 years ago
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What is the volume of 43.7 g of helium at stp?
tankabanditka [31]
Answer is: volume of helium is 244.72 liters.
m(He) = 43.7 g.
n(He) = m(He) ÷ M(He).
n(He) = 43.7 g ÷ 4 g/mol.
n(He) = 10.925 mol.
V(He) = n(He) · n(He).
V(He) = 10.925 mol · 22.4 L/mol.
V(He) = 244.72 L.
Vm - molar volume at STP.
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3 years ago
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Dilute solutions of NaHCO3 are sometimes used in treating acid burns.
Temka [501]

Answer:

We need 375 milliliters of 0.100 M NaHCO3 solution

Explanation:

Step 1: Data given

Initial molarity NaHCO3 = 0.100 M

Volume prepared solution = 750.0 mL

Molarity prepared solution = 0.05 M

Step 2: Calculate initial volume

C1*V1 = C2V2

⇒with C1 = the initial concentration = 0.100 M

⇒with V1 = The initial volume = TO BE DETERMINED

⇒with C2 = the new concentration = 0.0500M

⇒with v2 = the new volume = 750.0 mL = 0.750 L

0.100 M * V1 = 0.0500 M * 0.750 L

V1 = (0.0500M * 0.750L)/0.100 M

V1 = 0.375 L = 375 mL

We need 375 milliliters of 0.100 M NaHCO3 solution

8 0
3 years ago
An unknown compound contains only C , H , and O . Combustion of 5.90 g of this compound produced 11.8 g CO2 and 4.83 g H2O . Wha
Mamont248 [21]

Answer:

C₂H₄O

Explanation:

In a compound that contains Cabon, hydrogen and oxygen, the combustion produce CO₂ from the carbon, and H₂O from the hydrogens. Using the mass of the products we can solve the moles of Carbon and hydrogen. The empirical formula is the simplest whole-number of atoms present in a molecule.

<em>Moles CO₂ = Moles C:</em>

11.8g CO₂ * (1mol / 44g) = 0.268 moles CO₂ = 0.268 moles C * (12g/mol) =

3.216g C

<em>Moles H₂O = 1/2 moles H:</em>

4.83g H₂O * (1mol / 18g) = 0.268 moles H₂O * (2 mol H / 1 mol H₂O) =

0.537 mol H * (1g/mol) = 0.537g H

<em>Mass O to find moles O:</em>

5.90g Sample - 3.216g C - 0.537g H = 2.147g O * (1mol / 16g) = 0.134 moles O

<em>Ratio of atoms -Dividing in 0.134 moles-:</em>

C = 0.268mol C / 0.134 mol O = 2

H = 0.537mol H / 0.134 mol O = 4

O = 0.134mol O / 0.134 mol O = 1

Empirical formula is:

<h3>C₂H₄O</h3>

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3 years ago
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