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4 years ago

What is required in order to determine whether or not an object moves?

Chemistry

2 answers:

frez [133]4 years ago

4 0

Answer:

Explanation:

a reference point

AysviL [449]4 years ago

3 0

Answer:

A reference point.

Explanation:

Reference point:

A reference point is a point which is used to determine weather the object is in motion or not. The moving object is compared with the reference point.

An object is in the state of motion when its distance is changed relative to the other object or reference point.

In order to determine weather an object is moving or not a point of reference is needed.

Without reference point displacement can not be measured

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Amplitude measures the light's

charle [14.2K]

Answer:

d.) intensity

Explanation:

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3 years ago

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How many are molecules ( or formula) in each sample?

andre [41]

Answer:

$4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$
$16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

Explanation:

Number of molecules for $55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

$\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:$

Atomic mass of Na + H + C + 3(O) = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

$\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }$

$55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)$

$=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

Number of molecules for for $\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

$\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

$459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.$

$=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

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3 years ago

Which sample has more atoms in it? 150.0 g of gold or 10.0 g of lithium.

Deffense [45]

Answer:150g of gold

Explanation: There is a lot more gold and gold is significantly significantly more dense than lithium

6 0

3 years ago

How much time is needed to deposit 1.0 g of chromium metal from an aqueous solution of crcl3 using a current of 1.5 a?

PSYCHO15rus [73]

The metal component of the given compound, CrCl3, is chromium. The number of moles per 1 g of chromium is calculated through the equation below,

n = (1 g Cr)(1 mol Cr/51.996 g Cr)
n = 0.0192 mol Cr(3 electrons/1 mol Cr)
n = 0.0577 e-

Determine the number in charge by multiplying with Faraday's constant,

C = (0.0577 mol Cr)((1 F/1 mol e-)(96485 C/ 1F)
C = 5,566.87 C

Then, calculate time by dividing the charge with the current,

t = 5566.87 C/1.5 A
t = 3711.25 minutes
t = 61.84 hours

Answer: 61.84 hours

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3 years ago

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qaws [65]

im not sure but i think

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3 years ago