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3 years ago

Members of which group easily lose an electron to form a +1 cation?

Chemistry

2 answers:

castortr0y [4]3 years ago

4 0

alkali metal is the answer

Goryan [66]3 years ago

3 0

D...alkaline earth metals

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How many km are in 5.6mm? 5.6x103 5.6x10-6 5.6x10-3 5.6x106

il63 [147K]

Hey there :)

Question: How many km are in 5.6mm?

=>5.6x10^3

=>5.6x10^-6

=>5.6x10^-3

=> 5.6x10^6

Answer:-

$5.6 \times 10^{ - 6}$

Explanation:-

By using the formula-

$1millimeter = \frac{1}{1000000}$

As 1 with 6 zeros, we convert it into exponential form.

$= > \frac{1}{10^{6} }$

As this above value is fraction type, we can do the reciprocal, thus, the exponent gets a negative value.

$= > 10^{ - 6}$

Now combine with given question.

$= > 5.6 \times 10^{ - 6}$

7 0

2 years ago

A 50-gram sample has a half-life of 12 days. How much material will remain after 12 days?

Elis [28]

A 50-gram sample with a half-life of 12 days will have a remaining mass of 25 grams after its 12-day half-life. Every cycle of a half-life, the sample will lose half of its mass, so if the half-life, itself, is 12 days and the time period passing is 12 days, one half-life has passed and the material will be halved.

8 0

3 years ago

Read 2 more answers

1) Draw a particle picture of liquid lemonade turning into a solid popsicle BEFORE AFTER

Juli2301 [7.4K]

Explanation:

The particles are further apart in liquids than in solids so when the lemonade changes from a liquid to a solid, the particles would become closer.

5 0

3 years ago

If you have 5 unknowns and they were secondary alcohol aldehyde, carboxylic acid, amine, and alkane Put plan to identify the unk

Finger [1]

Answer:

the unknowns

Explanation:

7 0

3 years ago

1.How many mL of 0.523 M HBr are needed to dissolve 8.60 g of CaCO3?

prohojiy [21]

Answer:

The answer to your question is:

Explanation:

1.-

HBr = 0.523 M V = ?

CaCO3 = 8.6 g

2HBr(aq) + CaCO₃(s) ⇒ CaBr₂(aq) + H₂O(l) + CO₂(g)

MW CaCO₃ = 40 + 12 + 48 = 100 g

MW HBr = 80 + 1 = 81 g

Molarity = moles / volume

100 g of CaCO₃ ---------------- 1 mol

8.6 g ---------------- x

x = (8.6 x 1) / 100

x = 0.086 moles

2 moles of HBr ----------------- 1 mol of CaCO₃

x ----------------- 0.086 moles

x = (0.086 x 2) / 1 = 0.172 moles of HBr

Volume = moles / molarity

Volume = 0.172/ 0.523 = 0.323 l or 323 ml of HBr

2.-

V = ? ml NaOH 0.487 M

V = 101 ml of 0.628 M MnSO₄

MnSO₄(aq) + 2NaOH(aq) ⇒ Mn(OH)₂(s) + Na₂SO₄(aq)

MW MnSO₄ = 55 g

MW NaOH = NaOH = 40 g

Moles = Molarity x volume

Moles = (0.628) x (0.101)

Moles = 0.065 moles of MnSO₄

1 mol of MnSO₄ ------------------ 2 moles of NaOH

0.065 ----------------- x

x = (0.065x 2) / 1

x = 0.131 moles of NaOH

Volume = moles / molarity

Volume = 0.131 / 0.487

Volume = 0.268 l or 268 ml of NaOH

4 0

3 years ago