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Luda [366]
3 years ago
6

What is the percent composition of carbon in caffeine (C₈H₁₀N₄O₂)?

Chemistry
1 answer:
podryga [215]3 years ago
5 0

Caffeine has the following percent composition: carbon 49.48%, hydrogen 5.19%, oxygen 16.48% and nitrogen 28.85%. Its molecular weight is 194.19 g/mol.

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How many grams of nano3 remain in solution at 20 ∘c?
Amanda [17]
<span>the table say that at 20 degree celcius 88.0g of NANO3 will remain dissolved in 100 gm of H2O so at 20 degree celcius 80.0g of H20 will dissolve (88.0g)x(80g/100g)=70.4g of NANO3 so at 20 degree celcius 86.3g-70.4g= 15.9 gram of NANO3 will come out of solution.</span>
8 0
3 years ago
True or False
lozanna [386]
1. False (gravity)
2. True
3. False
4.True
5. False

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5 0
3 years ago
SOMEONE PLEASE HELP
Zinaida [17]

Answer:

6. 7870 kg/m³ (3 s.f.)

7. 33.4 g (3 s.f.)

8. 12600 kg/m³ (3 s.f.)

Explanation:

6. The SI unit for density is kg/m³. Thus convert the mass to Kg and volume to m³ first.

1 kg= 1000g

1m³= 1 ×10⁶ cm³

Mass of iron bar

= 64.2g

= 64.2 ÷1000 kg

= 0.0642 kg

Volume of iron bar

= 8.16 cm³

= 8.16 ÷ 10⁶

= 8.16 \times 10^{ - 6} \:  kg

\boxed{density =  \frac{mass}{volume} }

Density of iron bar

=  \frac{0.0642}{8.16 \times 10^{ - 6} }

= 7870 kg/m³ (3 s.f.)

7.

\boxed{mass = density \:  \times volume}

Mass

= 1.16 ×28.8

= 33.408 g

= 33.4 g (3 s.f.)

8. Volume of brick

= 12 cm³

= 12  \times  10^{ - 6}  \: m^{3}  \\  = 1.2 \times 10^{ - 5}  \: m ^{3}

Mass of brick

= 151 g

= 151 ÷ 1000 kg

= 0.151 kg

Density of brick

= mass ÷ volume

=  \frac{0.151} {1.25 \times 10^{ - 5} }  \\  = 12600 \: kg/ {m}^{3}

(3 s.f.)

6 0
3 years ago
The reactant concentration in a first-order reaction was 8.10×10−2 M M after 15.0 s s and 1.80×10−3 M M after 90.0 s s . What is
kipiarov [429]

Answer:

The answer to the question is

The rate constant for the reaction is 1.056×10⁻³ M/s

Explanation:

To solve the question, e note that

For a zero order reaction, the rate law is given by

[A] = -k×t + [A]₀

This can be represented by the linear equation y = mx + c

Such that y = [A], m which is the gradient is = -k, and the intercept c = [A]₀

Therefore the rate constant k which is the gradient is given by

Gradient = \frac{[A]_{2} - [A]_{1}  }{t_{2} - t_{1}  }  where [A]₁ = 8.10×10⁻² M and [A]₂ = 1.80×10⁻³ M

= \frac{1.80*10^{-3} M- 8.10*10^{-2} M}{90 s - 15 s} = -0.001056 M/s = -1.056×10⁻³ M/s

Threfore k = 1.056×10⁻³ M/s

3 0
3 years ago
What are the names of the components (each shown with a leader and a line) of a circuit shown in the diagram? Please help!!!!
alekssr [168]
Maybe: Battery, wire, lamp and switch
6 0
3 years ago
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