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2 years ago

Drag the images into the correct category. Contact Forces Noncontact Forces

Chemistry

1 answer:

Nitella [24]2 years ago

8 0

Answer:

Where are the categories, we need an image for reference or something to better understand the question

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How hot, in degrees Celsius, would the air inside the balloon have to get in order for the balloon to lift off the ground? Assum

SVETLANKA909090 [29]

Answer:

The temperature in degress Celsius is 52.25°C

Explanation:

According the equation:

$V(\rho _{a} -\rho _{t})g=mg\\\rho _{a} -\rho _{t}=\frac{n}{V} =\frac{340}{2950} =0.115kg/m^{3}$

$\rho _{t}=\rho _{a} -0.115=1.2-0.115=1.085kg/m^{3}$

The temperature is:

$T=\frac{P}{\rho _{t} R} =\frac{1.013x10^{5} }{1.085*287.05} =325.25K=52.25$ °C

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4 years ago

Help please ASAP 10 point

charle [14.2K]

6a. physical reaction
6b. endothermic reaction
7a. physical reaction
7b. the reactants are carbon dioxide, water and light energy. the product is starch
7c. exothermic reaction

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4 years ago

Identify the reaction type shown and describe the clues that were used to identify it.

Vika [28.1K]

The type of reaction of Fe2O3 + 2SiO2 → Fe2Si2O7 would be that it is a synthesis reaction. It is a type of chemical reaction in which two or more simple substances combine to form a more complex product. Hope this answers the question.

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3 years ago

Read 2 more answers

Reactioneaza total 22 kg calcar de puritate 80 % cu o solutie de acid clorhidric de concentratie 30 %. Stiind ca rezulta clorura

WINSTONCH [101]

Answer:

De la Alex;)(da alex ăla de e din cls cu tn:)

7 0

3 years ago

A student dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL . The student notices that the

Tresset [83]

Answer:

a. Molarity= $M =2.1x10^{-1}M$

b. Molality= $m=2.0x10^{-1}m$

Explanation:

Hello,

In this case, given the information about the aniline, whose molar mass is 93g/mol, one could assume the volume of the solution is just 200 mL (0.200 L) as no volume change is observed when mixing, therefore, the molarity results:

$M=\frac{n_{solute}}{V_{solution}} =\frac{3.9g*\frac{1mol}{93g} }{0.2L} =2.1x10^{-1}M$

Moreover, the molality:

$m=\frac{n_{solute}}{m_{solvent}} =\frac{3.9g*\frac{1mol}{93g} }{0.2L*\frac{1.05kg}{1L} } =2.0x10^{-1}m$

Best regards.

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3 years ago