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3 years ago

How many moles of oxygen are in 8.24 moles Mg(NO3)2

Chemistry

1 answer:

malfutka [58]3 years ago

5 0

Answer:

49.4 mol Oxygen

Explanation:

Mg(NO3)2 ----- 6 O

1 mol 6 mol

8.24 mol x mol

x = 8.24*6/1 = 49.44 mol ≈ 49.4 mol Oxygen

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which best describes an element? - a pure substance - a type of a mixture - a pure compound - an impure substance

Nikitich [7]

An element is a pure substance which means it cannot be broken down by any means.

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3 years ago

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A 3.00 g mass of compound x was added to 50.0 g of water and it is found that the freezing point has decreased by 1.25 degrees c

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Answer:

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Explanation:

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3 years ago

Balancing the reaction by oxidation number method k2cr2o7+sncl2+hcl

Mumz [18]

Answer:

$K_2Cr_2O_7 (aq) + 14 HCl (aq) + 3 SnCl_2 (aq)\rightarrow 2 CrCl_3 (aq) + 7 H_2O (l) + 3 SnCl_4 (aq) + 2 KCl (aq)$

Explanation:

The products of this reaction are given by:

$K_2Cr_2O_7 (aq) + SnCl_2 (aq) + HCl (aq)\rightarrow KCl (aq) + SnCl_4 (aq) + CrCl_3 (aq) + H_2O (l)$

Firstly, dichromate anion becomes chromium(III) cation, let's write this change:

$Cr_2O_7^{2-} (aq)\rightarrow Cr^{3+} (aq)$

The following steps should be taken:

balance the main element, chromium: multiply the right side by 2 to get 2 chromium species on both side:

$Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq)$

balance oxygen atoms by adding 7 water molecules on the right:

$Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

balance the hydrogen atoms by adding 14 protons on the left:

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

balance the charge (the total net charge on the left is 12+, on the right we have 6+, so 6 electrons are needed on the left):

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

Similarly, tin(II) cation becomes tin(IV) cation:

$Sn^{2+} (aq)\rightarrow Sn^{4+} (aq) + 2e^-$

Now that we have the two half-equations, multiply the second one by 3, so that it also has 6 electrons that will be cancelled out upon addition of the two half-equations:

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

$3 Sn^{2+} (aq)\rightarrow 3 Sn^{4+} (aq) + 6e^-$

Add them together:

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 3 Sn^{2+} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l) + 3 Sn^{4+} (aq)$

Adding the ions spectators:

$K_2Cr_2O_7 (aq) + 14 HCl (aq) + 3 SnCl_2 (aq)\rightarrow 2 CrCl_3 (aq) + 7 H_2O (l) + 3 SnCl_4 (aq) + 2 KCl (aq)$

7 0

3 years ago

Pls help me with my chemistry finals test.

Nutka1998 [239]

What do you need help with? Can you specify?

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2 years ago

If it requires 23.4 milliliters of 0.65 molar barium hydroxide to neutralize 42.5 milliliters of nitric acid, solve for the mola

rodikova [14]

Volume Ba(OH)2 = 23.4 mL in liters :

23.4 / 1000 => 0.0234 L

Molarity Ba(OH)2 = 0.65 M

Volume HNO3 = 42.5 mL in liters:

42.5 / 1000 => 0.0425 L

number of moles Ba(OH)2 :

n = M x V

n = 0.65 x 0.0234

n = 0.01521 moles of Ba(OH)2

Mole ratio :

<span>Ba(OH)2 + 2 HNO3 = Ba(NO3)2 + 2 H2O
</span>
1 mole Ba(OH)2 ---------------- 2 moles HNO3
0.01521 moles ----------------- moles HNO3

moles HNO3 = 0.01521 x 2 / 1

moles HNO3 = 0.03042 / 1

= 0.03042 moles HNO3

Therefore:

M ( HNO3 ) = n / volume ( HNO3 )

M ( HNO3 ) = 0.03042 / 0.0425

M ( HNO3 ) = 0.715 M

5 0

2 years ago