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3 years ago

TRUE OR FALSE 3D printing waste less material than traditional manufacturing.

Chemistry

1 answer:

jeka57 [31]3 years ago

3 0

True... there are better ways using less materials and printing is that

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A substance X contains 10 gram of calcium carbonate calculate the number of mole of calcium carbonate present in X

Tasya [4]

$\LARGE{ \boxed{ \rm{ \red{Required \: answer}}}}$

☃️ Chemical formulae ➝ $\sf{CaCO_3}$

For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.

$\boxed{ \sf{No. \: of \: moles = \frac{given \: weight}{molecular \: weight} }}$

<h3>Solution:</h3>

Atomic weight of elements:

Ca = 40

C = 12

O = 16

❍ Molecular weight of $\sf{CaCO_3}$

= 40 + 12 + 3 × 16

= 52 + 48

= 100 g/mol

❍ Given weight: 10 g

Then, no. of moles,

⇛ No. of moles = 10 g / 100 g mol‐¹

⇛ No. of moles = 0.1 moles

☄ No. of moles of Calcium carbonate in that substance = 0.1 moles

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3 0

3 years ago

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Students are designing an experiment to test the law of conservation of mass using the following materials: 10 g baking soda, 30

Ivan

it would be 50ml hope th

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2 years ago

A student has a cube with each side is 4 cm in length, what is the volume? What is the measurement unit? Select ALL that apply.

Margaret [11]

Answer:

64cm^3

Explanation:

To get the volume just multiple 4*4*4

4 0

2 years ago

What is the main function of the part labeled Y in the model?

Tanya [424]

Answer

Explanation

the answer is D

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2 years ago

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A 3.0-L sample of helium was placed in container fitted with a porous membrane. Half of the helium effused through the membrane

zhenek [66]

Explanation:

It is known that rate of effusion of gases are inversely proportional to the square root of their molar masses.

And, half of the helium (1.5 L) effused in 24 hour. So, the rate of effusion of He gas is calculated as follows.

$\frac{1.5 L}{24 hr}$

= 0.0625 L/hr

As, molar mass of He is 4 g/mol and molar mass of $O_{2}$ is 32 g/ mol.

Now,

$\frac{\text{Rate of He}}{\text{Rate of Oxygen}} = \sqrt{\frac{32}{4}$

= 2.83

or, rate of $O_{2} = \frac{\text{Rate of He}}{2.83}$

= $\frac{0.0625 L/hr}{2.83}$

Rate of $O_{2}$ = 0.022 L/hr.

This means that 0.022 L of $O_{2}$ gas effuses in 1 hr

So, time taken for the effusion of 1.5 L of $O_{2}$ gas is calculated as follows.

$\frac{1.5 L}{0.022 L/hr}$

= 68.18 hour

Thus, we can conclude that 68.18 hours will it take for half of the oxygen to effuse through the membrane.

3 0

3 years ago