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amm1812
3 years ago
6

Can someone please help me please?

Chemistry
2 answers:
NikAS [45]3 years ago
7 0

Answer:

i know it

Explanation:

it s D the answer bo

ro

mark me brainliest

Step2247 [10]3 years ago
4 0

Answer:

Part B

A reaction that has the same number and type of atoms on each side of the equation.

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What is the name of(NH4)3PO4
Tju [1.3M]

Answer:

What is the name of the compound (NH4) 3PO4? (NH)4+ is an ammonium radical and (PO4) 3- is a phospate radical. When we start writing the compound,the valency of phosphate goes to ammonium as subscript,and that of ammonium goes to phosphate. Hence the formula (NH4)3 PO4 and it's name is ammonium phospate.

Explanation:

5 0
3 years ago
Sulfurous acid, H2SO3, breaks down into water (H2O) and sulfur dioxide (SO2).
Anna35 [415]

The answer is,

<u>A. 1 atom of sulfur</u>

Please rate <u>Brainliest</u>   (:

3 0
4 years ago
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What is the molarity of a solution that contains 3.66 mol KCl in 7.98 L solution?
QveST [7]

Answer:

.4586M

Explanation:

It's simple, just devide the mol value by the volume value. So 3.66mol/7.98L=.4586M

4 0
3 years ago
What is the balanced form of the following equation?<br> Br2 + S2O32- + H2O → Br1- + SO42- + H+
Darina [25.2K]

Answer:

5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺

Explanation:

We will balance the redox reaction through the ion-electron method.

Step 1: Identify both half-reactions

Reduction: Br₂ ⇒ Br⁻

Oxidation: S₂O₃²⁻ ⇒ SO₄²⁻

Step 2: Perform the mass balance, adding H⁺ and H₂O where appropriate

Br₂ ⇒ 2 Br⁻

5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺

Step 3: Perform the charge balance, adding electrons where appropriate

2 e⁻ + Br₂ ⇒ 2 Br⁻

5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻

Step 4: Make the number of electrons gained and lost equal

5 × (2 e⁻ + Br₂ ⇒ 2 Br⁻)

1 × (5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻)

Step 5: Add both half-reactions

5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺

8 0
3 years ago
In the reaction MgCI2+ 2KOH —&gt; Mg(OH)2 + 2KCI if 6 moles MgCI2 are added to 6 miles KOH which is the limiting reagent
Ratling [72]

KOH is the limiting reagent

The molar ratio is 1:2

So if you have 6 moles of MgCl2 you need 12 moles of KOH

7 0
4 years ago
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