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devlian [24]
2 years ago
14

How many moles are in 150 grams lf NaOH

Chemistry
1 answer:
ddd [48]2 years ago
7 0

Answer:

3.75 mole

Explanation:

molecular mass of NaOH = 39.97 gm

i.e,

==>39.97 gm NaOH contains 1 mole of NaOH

==>1 gm of NaOH contains 1/39.97 mole of NaOH

==>150 gm of NaOH contains (1/39.97) *150 mole of NaOH

==>150 gm of NaOH contains 3.75 mole of NaOH.

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How many sodium atoms does 2NaOH have?
attashe74 [19]

Answer:

1

Explanation:

5 0
3 years ago
Read 2 more answers
A student made the Lewis dot diagram of a compound as shown. Mg is written with two dots shown on its top. Cl is written on the
blsea [12.9K]

Answer:

The dots were not properly located and arrows are not used in Lewis structures

Explanation:

If we intend to write a Lewis structure for a compound, that lewis structure must consist of only dots. These dots actually show the valence electrons on the outermost shell of the molecule.

We do not involve arrows when writing dot electron structures for compounds. The valence electrons of magnesium ought not to be written together because they are not a lone pair, rather they are two unpaired electrons. The use of an arrow suggests a coordinate covalent bond which is not the case here.

The correct lewis structure for MgCl2 is shown in the image attached to this answer.

7 0
3 years ago
An herbicide contains only C, H, Cl, and N. The complete combustion of a 200.0 mg sample of the herbicide in excess oxygen produ
MAVERICK [17]

Answer:

%C = 56,1%

%H = 5,5%

%Cl = 27,6%

%N = 10,8%

Explanation:

The moles of CO₂ are the same than moles of C in the herbicide.

Moles of H₂O are ¹/₂ of moles of H in the herbicide.

Moles of CO₂ are obtained using:

n = PV/RT

Where, in STP: P is 1 atm; V is 0,2092L; R is 0,082atmL/molK; T is 273 K

moles of CO₂ are: 9,345x10⁻³ mol≡ mol of C×12,01g/mol = <em>0,1122 g of C ≡ 112,2mg of C</em>

In the same way, moles of H₂O are 5,450x10⁻³mol×2 =0,1090 mol of H×1,01g/mol = <em>0,0110 g of H ≡ 11,0mg of H</em>

As you have 55,14 mg of Cl, the mg of N are:

200,0mg - 112,2 mg of C - 11,0 mg of H - 55,14 mg of Cl = 21,66 mg of N

Thus, precent composition of the herbicide is:

%C = \frac{112,2 mgC}{200,0mg}×100 = 56,1%C

%H = \frac{11,0 mgH}{200,0mg}×100 = 5,5%H

%Cl = \frac{55,14 mgCl}{200,0mg}×100 = 27,6%Cl

%N = \frac{21,66 mgN}{200,0mg}×100 = 10,8%N

I hope it helps!

3 0
3 years ago
How much of a 400g sample remains after 4 years if a radioactive isotope has a half-life of 2 years?
timama [110]

Answer:

100 g

Explanation:

From the question given above, the following data were obtained:

Original amount (N₀) = 400 g

Time (t) = 4 years

Half-life (t½) = 2 years

Amount remaining (N) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Time (t) = 4 years

Half-life (t½) = 2 years

Number of half-lives (n) =?

n = t / t½

n = 4 / 2

n = 2

Thus, 2 half-lives has elapsed.

Finally, we shall determine the amount remaining of the radioactive isotope. This can be obtained as follow:

Original amount (N₀) = 400 g

Number of half-lives (n) = 2

Amount remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2² × 400

N = 1/4 × 400

N = 0.25 × 400

N = 100 g

Thus, the amount of the radioactive isotope remaing is the 100 g.

3 0
3 years ago
78.9 + 890.43 - 21 = 9.5 x 10^2
maria [59]

948 or 9.48 x 10^2

There are two sets of rules for significant figures

• One set for addition and subtraction

• Another set for multiplication and division

You used the set for multiplication and division.

This problem involves addition and subtraction, and the rule is

The number of places after the decimal point in the answer must be <em>no greater than the number of decimal places in every term</em> in the sum.

Thus, we have

78.9

+890.43

-21.

= 948.33

The "21" term has the fewest digits after the decimal point (none), so the answer must have no digits after the decimal point.

To the correct answer is 948 = 9.48 x 10^2. It has three significant figures.

8 0
3 years ago
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