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Karo-lina-s [1.5K]
3 years ago
14

In the second step of this reaction, isocyanic acid reacts to form melamine and carbon dioxide: HNCO(l)→C3N3(NH2)3(l)+CO2(g) Bal

ance this reaction and enter the coefficients below.
Chemistry
1 answer:
kykrilka [37]3 years ago
8 0

Answer:

The coefficients are 6, 1, 3

Explanation:

HNCO →C3N3(NH2)3 + CO2

From the above equation, there are a total of 6 atoms of nitrogen on the right side and 1atom on the left. It can be balance by putting 6 in front of HNCO as shown below:

6HNCO → C3N3(NH2)3 + CO2

Now there are 6 atoms of carbon on the left side and 4 atoms on the right side. It can be balance by putting 3 in front of CO2 as shown below:

6HNCO → C3N3(NH2)3 + 3CO2

Now the equation is balanced as the numbers of atoms of the different elements on both sides of the equation are the same.

The coefficients are 6, 1, 3

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What is the half life of the element in the picture<br><br> HELP BRAINLIEST
stira [4]

Answer:

6 days

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 100 mg

Amount remaining (N) = 6. 25 mg

Time (t) = 24 days

Half life (t½) =?

Next, we shall determine the decay constant. This can be obtained as follow:

Original amount (N₀) = 100 mg

Amount remaining (N) = 6. 25 mg

Time (t) = 24 days

Decay constant (K) =?

Log (N₀/N) = kt / 2.303

Log (100/6.25) = k × 24 / 2.303

Log 16 = k × 24 / 2.303

1.2041 = k × 24 / 2.303

Cross multiply

k × 24 = 1.2041 × 2.303

Divide both side by 24

K = (1.2041 × 2.303) / 24

K = 0.1155 /day

Finally, we shall determine the half-life of the isotope as follow:

Decay constant (K) = 0.1155 /day

Half life (t½) =?

t½ = 0.693 / K

t½ = 0.693 / 0.1155

t½ = 6 days

Therefore, the half-life of the isotope is 6 days

5 0
2 years ago
You are asked to identify compound X (a white, crystalline solid), which was extracted from a plant seized by customs inspectors
igor_vitrenko [27]

Answer:

C2H2O4

Explanation:

To get the molecular formula, we first get the empirical formula. This can be done by dividing the percentage compositions by the atomic masses. The percentage compositions are shown as follows :

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H = 2.239%

O = 100 - ( 26.86 + 2.239) = 70.901%

We then proceed to divide by their atomic masses. Atomic mass of carbon is 12 a.m.u , H = 1 a.m.u , O = 16 a.m.u

The division is as follows:

C = 26.86/12 = 2.2383

H = 2.239/1 = 2.239

O = 70.901/16 = 4.4313

We now divide each by the smallest number I.e 2.2383

C = 2.2383/2.2383 = 1

H = 2.239/2.2383 = 1

O = 4.4313/2.2383 = 1.98 = 2

Thus, the empirical formula is CHO2.

To get the molecular formula, we use the molar mass .

(CHO2)n = 90

We add the atomic masses multiplied by n.

(12 + 1 + 2(16))n = 90

45n = 90

n = 90/45 = 2.

Thus , the molecular formula is C2H2O4

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3 years ago
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Answer:

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