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IrinaK [193]
3 years ago
5

Which element has similar properties to lithium? Explain your reasoning:

Chemistry
1 answer:
Sphinxa [80]3 years ago
4 0

Answer:

sodium/potassium/rubidium/caesium/francium

Explanation:

all are group I elements, so they all have similar properties

You might be interested in
7. A solution containing 90grams of KNO3 per 100. grams of H2O at
Tomtit [17]

Answer:

(4) concentrated and supersaturated

Explanation:

At 50.°C, 90g of KNO3 lies above the solubility curve [on the Regents Reference Table G]. This indicates that the solution is supersaturated, meaning it contains more solute than will naturally dissolve, and was formed when a saturated solution cooled. Furthermore, the percent concentration of this solution is 90% KNO3 making this solution concentrated. This can be calculated using the formula for mass percent concentration.

Percent Mass = <u>Mass of Solute (g)</u> x 100

                         Mass of Solution (g)

5 0
3 years ago
What In the formula Na3PO4 how many moles of sodium? How many moles of phosphorus? How many moles of oxygen?
yulyashka [42]
Sodium is Na, so there's 3 moles of that. Phosphorus is P, there's 1 mole of that. Oxygen is O, there's 4 moles of that.
5 0
3 years ago
What is the voltage for the following cell: Cu(s)| Cu+(aq) || Mg2+(aq) |Mg(s)?
Julli [10]

Answer:

I think it is number 4 which is 3.87 v

6 0
3 years ago
What are the signs of the enthalpy change (ΔH°) and the entropy change (ΔS°) for the condensation of CS2(g)?
VARVARA [1.3K]

Answer:

∆H is negative

∆S is negative

Explanation:

The condensation of CS2 implies a phase change from gaseous state to liquid state. The energy of the gaseous particles is greater than that of the liquid particles hence energy is given out when a substance changes from gaseous state to liquid state hence the process is exothermic and ∆H is negative.

Changing from gaseous state to liquid states leads to a decrease in entropy hence ∆S is negative. Liquid particles are more orderly than particles of a gas.

5 0
4 years ago
Rank and explain how the freezing point of 0.100 m solutions of the following ionic electrolytes compare. List from lowest freez
Elden [556K]

Answer:

The solutions are listed from the lowest freezing point to highest freezing point.

Mg₃(PO₄)₂ -  AlBr₃ - BeBr₂ -  KBr

Explanation:

Colligative property of freezing point, is the freezing point depression.

Formula is:

ΔT = Kf . m . i

Where ΔT = T° freezing pure solvent - T° freezing solution

Kf is the cryoscopic constant

m is molality and i is the Van't Hoff factor (number of ions, dissolved in solution)

In this case, T° freezing pure solvent is 0° because it's water, so the Kf is 1.86 °C/m, and m is 0.1 molal.

The i modifies the T° freezing solution. The four salts, are ionic compounds, so for each case:

BeBr₂  → Be²⁺  +  2Br⁻   i = 3

AlBr₃ →  Al³⁺  + 3Br⁻   i = 4

Mg₃(PO₄)₂  →  3Mg²⁺  +  2PO₄⁻³    i = 5

KBr →  K⁺  +  Br⁻   i = 2

Solution of Mg₃(PO₄)₂ will have the lowest temperature of freezing and the solution of KBr, the highest (always lower, than 0°).

8 0
3 years ago
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