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3 years ago

In what orbital would to valence electrons of strontium be found? a.) 5s b.) 4s c.) none d.) 5p

Chemistry

1 answer:

KonstantinChe [14]3 years ago

4 0

Answer:

The correct answer is 5s

Explanation:

Strontium atomic number = 38

Electronic configuration is $$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 5s^{2}$

As $5s$ orbital is valence orbital in strontium, the outer electrons exists in that orbital only.

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As more discovers were made, whittaker proposed classifying orginsims into five _____

guapka [62]

Answer:

As more discovers were made, Whittaker proposed classifying organisms into five Kingdoms

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3 years ago

Starting from a 1 M stock of NaCl, how will you make 50 ml of 0.15 M NaCl?

WARRIOR [948]

Answer:

We need 7.5 mL of the 1M stock of NaCl

Explanation:

Data given:

Stock = 1M this means 1 mol/ L

A 0.15 M solution of 50 mL has 0.0075 moles NaCl per 50 mL

Step 2: Calculate the volume of stock we need

The moles of solute will be constant

and n = M*V

M1*V1 = M2*V2

⇒ with M1 = the initial molair concentration = 1M

⇒ with V1 = the volume we need of the stock

⇒ with V2 = the volume we want to make of the new solution = 50 mL = 0.05 L

⇒ with M2 = the concentration of the new solution = 0.15 M

1*V1 = 0.15*(50)

V1 = 7.5 mL

Since 0.0075 L of 1M solution contains 0.0075 moles

50 mL solution will contain also 0.0075 moles but will have a molair concentration of 0.0075 moles / 0.05 L =0.15 M

We need 7.5 mL of the 1M stock of NaCl

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3 years ago

There are two groups of waves with the same amplitude. One contains waves of short wavelength and the other has waves of long wa

jolli1 [7]

Answer:

i think the long wavelength has more energy

Explanation:

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3 years ago

Read 2 more answers

An acid with molar mass 84.48 g/mol is titrated with 0.650 M KOH. What volume of KOH solution is needed to titrate 1.70 grams of

RSB [31]

Answer:

$V=0.0310L=3.10mL$

Explanation:

Hello.

In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:

$n_{acid}=n_{KOH}$

Thus, since we are given 1.70 g of the acid, we compute the moles of acid that were titrated:

$n_{acid}=1.70g*\frac{1mol}{84.48g}=0.0201mol$

Which equal the moles of KOH. In such a way, since the molarity is defined as moles over liters (M=n/V), the liters are moles over molarity (V=n/M), thus, the resulting volume is:

$V=\frac{0.0201mol}{0.650mol/L}\\\\V=0.0310L=3.10mL$