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2 years ago

When baking soda is mixed with lemon juice, bubbles are formed with the evolution of a gas. What type of change is it? Explain.

Physics

1 answer:

lorasvet [3.4K]2 years ago

5 0

$\large{\underline{\underline{\pmb{\frak {\color {red}{Question:}}}}}}$

$\sf \red{When \: baking \: soda \: is \: mixed \: with \: lemon \: juice, \: bubbles \: are \: formed \: with \: the \: evolution \: of \: a \: gas.}$

$\large{\underline{\underline{\pmb{\frak {\color {blue}{Answer:}}}}}}$

When baking soda is mixed with lemon juice, bubbles are formed with the evolution of a gas. The gas is formed in the reaction is Carbon dioxide. $CO_{2}$ is formed.

The change which happened in this reaction is a chemical change.

$\boxed{ \frak \green{Explanation}}$

Since, in chemical change we can't bring a substance to it's actual form how it was in earlier.

Examples: burning of paper is chemical, since we can't get the fine paper again after it is burnt.

Thus, the above reaction is also a chemical change, since we can't get back the lemon juice how it was earlier.

$\boxed{ \frak \red{Brainlysamurai}}$

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2 years ago

A vertical spring with a spring constant of 310 N/m is mounted on the floor. From directly above the spring, which is unstrained

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3 years ago

PLEASE SOLVE QUICKLY!!!<br> Solve for A, B, and C from graph

hram777 [196]

A = 59.35cm

B = 196.56g

C = 74.65g

<u>Explanation:</u>

We know,

$x = \frac{L}{\frac{W}{F} +1}$

and L = x+y

Total length, L = 100cm

Weight of Beam, W = 71.8g

Center of mass, x = 49.2cm

Added weight, F = 240g

Position weight placed from fulcrum, y = ?

$L-y = \frac{L}{\frac{W}{F}+1 } \\100 - y = \frac{100}{\frac{71.8}{49.2}+1 } \\100 - y = \frac{100}{1.46+1}\\\\100 - y = \frac{100}{2.46} \\100-y = 40.65\\\\y = 59.35cm$

Therefore, position weight placed from fulcrum is 59.35cm

Total length, L = 100cm

Center of mass, x = 47.8 cm

Added weight, F = 180g

Position weight placed from fulcrum, y = 12.4cm

Weight of Beam, W = ?

$47.8 = \frac{100}{\frac{W}{180}+1 }\\\47.8 = \frac{100}{\frac{W+180}{180} } \\\\47.8 = \frac{100 X 180}{W+180}\\ \\47.8W + 47.8 X 180 = 18000\\47.8W = 18000 - 8604\\W = \frac{9396}{47.8}\\ W = 196.56g$

Therefore, weight of the beam is 196.56g

Total length, L = 100cm

Center of mass, x = 50.8 cm

Position weight placed from fulcrum, y = 9.8cm

Weight of Beam, W = 72.3g

Added weight, F = ?

$50.8 = \frac{100}{\frac{72.3}{F}+1 }\\\ 50.8 = \frac{100}{\frac{72.3+F}{F} } \\\\50.8 = \frac{100 X F}{72.3+F}\\ \\50.8 X 72.3 + 50.8 X F = 100F\\\\3672.84 = 100F-50.8F\\3672.84 = 49.2F\\F = 74.65g$

Therefore, Added weight F is 74.65g

A = 59.35cm

B = 196.56g

C = 74.65g

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