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2 years ago

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On combustion of 0.2 gm of organic compound gave 0.147g of CO2, 0.12g of H2O and 74.6 ml of nitrogen gas at STP. Find emperical

formula of compound.?

1 answer:

zhuklara [117]2 years ago

5 0

Answer:

0.2 is conpound Co2 STP.

Explanation:

on combustion of 0.2 gm of organic compound gave 0.147g of CO2, 0.12g of H2O and 74.6 ml of nitrogen gas at STP. Find emperical formula of compound.?

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Practice Problem: True Stress and Strain A cylindrical specimen of a metal alloy 49.7 mm long and 9.72 mm in diameter is stresse

amm1812

Answer:

The true stress required = 379 MPa

Explanation:

True Stress is the ratio of the internal resistive force to the instantaneous cross-sectional area of the specimen. True Strain is the natural log to the extended length after which load applied to the original length. The cold working stress – strain curve relation is as follows,

σ(t) = K (ε(t))ⁿ, σ(t) is the true stress, ε(t) is the true strain, K is the strength coefficient and n is the strain hardening exponent

True strain is given by

Epsilon t =㏑ (l/l₀)

Substitute㏑(l/l₀) for ε(t)

σ(t) = K(㏑(l/l₀))ⁿ

Given values l₀ = 49.7mm, l =51.7mm , n =0.2 , σ(t) =379Mpa

379 x 10⁶ = K (㏑(51.7/49.7))^0.2

K = 379 x 10⁶/(㏑(51.7/49.7))^0.2

K = 723.48 MPa

Knowing the constant value would be same as the same material is being used in the second test, we can find out the true stress using the above formula replacing the value of the constant.

σ(t) = K(㏑(l/l₀))ⁿ

l₀ = 49.7mm, l = 51.7mm, n = 0.2, K = 723.48Mpa

σ(t) = 723.48 x 106 x (㏑(51.7/49.7))^0.2

σ(t) = 379 MPa

The true stress necessary to plastically elongate the specimen is 379 MPa.

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3 years ago

Read 2 more answers

Standard temperature and pressure are _____.

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the standard pressure is 1atm.
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The _______ force between each planet and the Sun keeps the planets in orbit around the Sun.

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Name two elements in the modern periodic table that break Mendeléey's rule that the elements should be arranged in order of rela

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X, Y & Z represents 2, 12 & 17 relatively a. Write the electronic configuration of x, y, and z b. Place the elements in

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The electronic configuration of x, y, and z.

$x=1s^2$

$Y= 1s^22s^22p^63s^2$

$Z= 1s^22s^22p^63s^23p^5$

<h3>What is an electronic configuration?</h3>

Electronic configuration, also called electronic structure, is the arrangement of electrons in energy levels around an atomic nucleus.

a.

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$Y= 1s^22s^22p^63s^2$

$Z= 1s^22s^22p^63s^23p^5$

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Y=Element is in period 3 and Group 2

Z=Element is in period 3 and group 17.

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