What coefficient values will balance the reaction shown? CS2 + Cl2? CCl4 + S2Cl2 A. 1,1,1,1 B. 1,3,1,1 C. 2,1,1,1 D. 1,2,1,2

How many moles are in 3.00x 10^22 atoms of He?

sp2606 [1]

Answer:

<h2>0.05 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{3.0 \times {10}^{22} }{6.02 \times {10}^{23} } \\ = 0.049833...$

We have the final answer as

<h3>0.05 moles</h3>

Hope this helps you

3 0

3 years ago

you have been observing an insect that defends itself from enemies by secreting a caustic liquid. analysis of the liquid shows i

Alexus [3.1K]

pH of the buffer solution is 1.76.

Chemical dissociation of formic acid in the water:

HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq)

The solution of formic acid and formate ions is a buffer.

[HCOO⁻] = 0.015 M; equilibrium concentration of formate ions

[HCOOH] + [HCOO⁻] = 1.45 M; sum of concentration of formic acid and formate

[HCOOH] = 1.45 M - 0.015 M

[HCOOH] = 1.435 M; equilibrium concentration of formic acid

pKa = -logKa

pKa = -log 1.8×10⁻⁴ M

pKa = 3.74

Henderson–Hasselbalch equation: pH = pKa + log(cs/ck)

pH = 3.74 + log (0.015 M/1.435 M)

pH = 3.74 - 1.98

pH = 1.76

More about buffer: brainly.com/question/4177791

#SPJ4

4 0

1 year ago

A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out

scZoUnD [109]

<u>Answer:</u> The mass percent of potassium bromide in the mixture is 9.996%

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For lead (II) bromide:</u>

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

$\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol$

The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

$2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3$

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = $\frac{2}{1}\times 0.0021=0.0042mol$ of potassium bromide

Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

$0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g$

To calculate the percentage composition of KBr in the mixture, we use the equation:

$\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100$

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

$\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%$

Hence, the percent by mass of KBr in the mixture is 9.996 %

5 0

3 years ago

A tank holds 6.90 m3 of water. What is the volume in ft3? (3.28 ft = 1 m, Remember significant figures, but this answer does not

AleksAgata [21]

Volume of tank = $6.9 m^{3}$ (given)

Since, $1 m = 3.28 ft$

So,

$1 m^{3} = 35.287 ft^{3}$

For $6.9 m^{3}$ :

$6.9\times 35.287 ft^{3} = 243.4803 ft^{3}$

The significant rule for multiplication, states that the number of significant figures in the answer obtained by multiplication is determined by the value with the lowest number of significant digits.

Since, the minimum number of decimal places in the above multiplication operation is 1 so, the final result must be upto 1 decimal place only.

$243.4803 ft^{3} \simeq 243.5 ft^{3}$

Hence, volume in $ft^{3}$ is 243.5.

5 1

2 years ago

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Graciella divided her grapes equally among 6 friends. If each friend received 16 grapes how many grapes did Graciella have?

DIA [1.3K]

Answer: She had 96 grapes.

Explanation:

3 0

2 years ago

Read 2 more answers