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2 years ago

Im posting seperate questions so just answer this one

Chemistry

1 answer:

vredina [299]2 years ago

6 0

On the lab the text is kind of too far zoomed out so u can’t really read it it’s like blurry

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Hii pls help me to balance the equation thanksss

Vlada [557]

Answer:

3Fe(s)+2O2(g)---->Fe3O4

this way you will have 3irons on both sides and 4 oxygens.

I hope this helps

3 0

3 years ago

Read 2 more answers

What is displacement reaction

jenyasd209 [6]

A displacement reaction (also known as a replacement reaction) is when one element is replaced by another compound. Ex) Fe+CuSO4=FeSO+Cu

3 0

3 years ago

Read 2 more answers

Radioactive decay can be described by the following equation where is the original amount of the substance, is the amount of the

soldi70 [24.7K]

Answer:

Iron remains = 17.49 mg

Explanation:

Half life of iron -55 = 2.737 years (Source)

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{2.737}\ year^{-1}$

The rate constant, k = 0.2533 year⁻¹

Time = 2.41 years

$[A_0]$ = 32.2 mg

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

So,

$[A_t]=32.2\times e^{-0.2533\times 2.41}\ mg$

$[A_t]=32.2\times e^{-0.610453}\ mg$

$[A_t]=17.49\ mg$

<u>Iron remains = 17.49 mg</u>

8 0

3 years ago

A sample of pure tin metal is dissolved in nitric acid to produce 15.00 mL of solution containing Sn2+. When this tin solution i

Rudik [331]

Answer:

1.00 M

Explanation:

Sn^2+ reacts with KMNO4 as follows;

5Sn^2+(aq) + 2MnO4^-(aq) + 16H^+(aq) ----> 5Sn^4+(aq) + 2Mn^+(aq) + 8H2O(l)

The number of moles of MnO4^- reacted = 42.1/1000 L × 0.145 mol/L

= 0.0061 moles

If 5 moles of Sn^2+ reacts with 2 moles of MnO4^-

x moles of Sn^2+ reacts with 0.0061 moles of MnO4^-

x= 5 × 0.0061/2

x= 0.015 moles

Since the volume of the Sn^2+ solution is 15.00mL or 0.015 L

number of moles = concentration × volume

Concentration = number of moles/volume

Concentration= 0.015 moles/0.015 L

Concentration = 1 M

6 0

3 years ago

Name the two possible products in the precipitation reaction of copper (II) chloride and sodium phosphate. Use the charges on th

satela [25.4K]

Answer:

General equation for a double-displacement reaction:

AB + CD --> AC + BD

• sodium chloride – NaCl copper sulfate – CuSO₄

NaCl + CuSO₄ --> Na₂SO₄ + CuCl₂

The products formed are sodium sulfate and copper (II) chloride.

Copper (II) chloride forms a blue colored solution.

• sodium hydroxide – NaOH copper sulfate – CuSO₄

NaOH + CuSO₄ --> Na₂SO₄ + Cu(OH)₂

The products formed are sodium sulfate and copper (II) hydroxide.

Copper (II) hydroxide forms a blue colored solution.

• sodium phosphate – Na₂HPO₂ copper sulfate – CuSO₄

Na₂HPO₄ + CuSO₄ --> Na₂SO₄ + CuHPO₄

The products formed are sodium sulfate and copper (II) hydrogen phosphate.

Copper (II) hydrogen phosphate forms a blue colored solution.

• sodium chloride – NaCl silver nitrate – AgNO₃

NaCl + AgNO₃--> AgCl + NaNO₃

The products formed are silver chloride and sodium nitrate.

Silver chloride forms a white precipitate.

• sodium hydroxide – NaOH silver nitrate – AgNO₃

NaOH + AgNO₃ --> NaNO₃ + AgOH

The products formed are silver hydroxide and sodium nitrate.

Silver hydroxide forms a white precipitate.

• sodium phosphate – Na₂HPO₄ silver nitrate – AgNO₃

Na₂HPO₄ + AgNO₃ --> NaNO₃ + Ag₂HPO₄

The products formed are sodium nitrate and silver hydrogen phosphate.

Silver hydrogen phosphate forms a colorless solution.

Explanation:

5 0

3 years ago