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tamaranim1 [39]
3 years ago
7

Which tool or tools would you use to measure and compare the mass of a cup of fresh water and a cup of salt water

Chemistry
1 answer:
Artemon [7]3 years ago
4 0
You could use a scale to measure the mass as well as a cup to hold the water. If you were comparing the two, you should also probably use a graduated cylinder to get the same amount of each type of water.

Hope this helped ^_^
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Fe3+(aq) (yellow) + SCN-(aq) (colorless) FeSCN2+(aq) (blood-red) Chloride ions are colorless. Potassium ions are also colorless.
harina [27]

Explanation:

Fe^{3+}\text{(aq)(yellow)}+SCN^-\text{(aq)(colorless)}\rightleftharpoons FeSCN^{2+}\text{(aq)(blood-red)}

K^+\text{(aq)(colorless)}+Cl^-\text{(aq)(colorless)}\rightarrow KCl

The above two reactions xcan also be written in form of single chemcial equation:

FeCl_3+K(SCN)\rightleftharpoons KCl+[Fe(SCN)]Cl_2(blood-red)

1. Color of ferric chloride solution is yellow. This is due to presence of ferric ions which have yellow color in their aqueous solutions.

2. KSCN has the colorless solution. This due to potassium ion forms colorless aqueous solution.

3. On mixing, KSCN with FeCl_3 we will get blood red color solution of [FeSCN]Cl_2.

5 0
3 years ago
Potassium + water → potassium hydroxide<br> +<br> hydrogen<br> formula
kvasek [131]

Answer:

If you help I'll help you deal?

3 0
2 years ago
Give the electron configuration for the following atoms using appropriate noble gas inner core abbreviation: Bi Cr Sr P 2. Give
malfutka [58]

Answer:

See explanation

Explanation:

The noble gas core electron configuration involves writing the inert gas core of an atom followed by the valence electrons. This is shown for the following atoms;

Bismuth;

[Xe]4f14 5d10 6s2 6p3

Chromium;

[Ar]4s1 3d5

Strontium;

[Kr]5s2

Phosphorus;

[Ne]3s2 3p3

2.

Bi

6p- n=6, l= 1, ml= 1, ms= 1/2

Cr

3d- n=3, l=2, ml=2,ms=1/2

Sr

5s- n=5, l=0, ml=0, ms=1/2

P

3p- n=3, l= 1, ml= 1, ms=1/2

3.

a) Tin (Sn) - [Kr] 5s2 4d10 5p2

b) Caesium (Cs)- [Xe] 6s1

c) Copper (Cu)- [Ar] 4s1 3d10

4 0
3 years ago
Consider two solutions, the first being 50.0 mL of 1.00 M CuSO4 and the second 50.0 mL of 2.00 M KOH. When the two solutions are
kolbaska11 [484]

Explanation:

Molarity of copper sulfate solution = 1.00 M

Volume of the copper sulfate solution  = 50.0 mL = 0.050 L

Moles of copper sulfate = n

1.00M=\frac{n}{0.050 L}

n = 0.050 L × 1.00 M= 0.050 mol

1 mol of copper sulfate has 1 mol of copper . Then 0.050 mol of copper sulfate has :

1\times 0.050 mol=0.050 mol of copper

a) Mass of 0.050 moles of copper = 0.050 mol × 63.5 g/mol =3.175 g

b) The identity of the compound which formed after the reaction is copper hydroxide.

c) The complete equation for the reaction that occurs when copper sulfate and potassium hydroxide are mixed:

CuSO_4(aq)+2KOH (aq)\rightarrow Cu(OH)_2(s)+K_2SO_4 (aq)

d) CuSO_4(aq)\rightarrow Cu^{2+}(aq) +SO_{4}^{2-}(aq)..[1]

KOH (aq)\rightarrow 2K^+(aq) +OH^-(aq)..[2]

Cu^{2+}(aq) +SO_{4}^{2-}(aq)+2K^+(aq) +2OH^-(aq)\rightarrow Cu(OH)_2(s)+SO_{4}^{2-}(aq)+2K^+(aq)

Common ion both sides are removed. The net ionic equation is given as:

Cu^{2+}(aq) +2OH^-(aq)\rightarrow Cu(OH)_2(s)(aq)

e) Volume of solution after mixing of both solution,V= 50 mL + 50ml = 100 mL

Mass of final solution ,m= 1 mL

Density of solution ,d= 1 g/mL (same as pure water)

m=d\time V=1 g/ml\times 100 mL = 100 g

Heat capacity of the solution = c = 4.186 J/g°C (same as pure water)

Change in temperature of the solution,ΔT = 27.7 °C- 21.5 °C=6.2°C

Q=mc\Delta T

Q=100 g\times 4.186 J/g ^oC\times 6.2^oC=2595.32 J=2.595 kJ

Enthalpy of the reaction = ΔH = \frac{Q}{\text{Moles of copper}}

ΔH = \frac{2.595 kJ}{0.050 mol}=51.90 kJ/mol

The ΔH for the reaction that occurs on mixing is 51.90 kJ/mol.

7 0
3 years ago
A 1.0 kg bottle of sodium carbonate (Na2CO3, 106.0 g/mol) is available to clean up 5.00 liters of spilled concentrated aqueous h
velikii [3]

Answer:

The correct answer is option 4, that is, there is exactly enough sodium carbonate.

Explanation:

Based on the given question, the reaction will be,

2 HCl (aq) + Na2CO3 (s) ⇒ 2 NaCl (aq) + CO2 (g) + H2O (l)

Therefore, for neutralizing 2 moles of HCl, one mole of Na2CO3 is required.

No of moles present in 1 Kg or 1000 grams of Na2CO3 will be,

Moles = Weight/Molecular mass of Na2CO3

Moles = 1000 / 106 = 9.43

Thus, 9.43 moles of Na2CO3 is present.

No of moles present in 1 liter of 9.75 M HCl is 9.75.

No. of moles present in 5 Liters of HCl (9.75 M),

= 5 × 9.75 = 48.75

Thus, for 2 moles of HCl 1 mole of Na2CO3 is required. Now for 48.75 moles of HCl, the moles required of Na2CO3 is 9.75. Therefore, for complete neutralization, the moles of Na2CO3 required is 9.75, and the present moles is 9.43.

Hence, there is exactly enough sodium carbonate.

8 0
3 years ago
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