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3 years ago

In the compound CaCO3 at the end of the compound represents the number of

Physics

1 answer:

Afina-wow [57]3 years ago

5 0

Answer:

Calculate the mass of 6.022 × 1023 molecule of Calcium carbonate (CaCO3).

Solution —

Molar mass (Molecular mass in gram) of CaCO3 = 40+12+3×16 = 100 g

No. of moles of CaCO3

= No. of molecules/Avogadro constant

= 6.022 × 1023/ 6.022 × 1023

= 1 mole

Mass of CaCO3

= No. of moles × molar mass

= 1 × 100 g = 100 g.

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A sample of ascorbic acid (vitamin C) is synthesized in the laboratory. It contains 1.50 g of carbon and 2.00 g of oxygen. Anoth

Karo-lina-s [1.5K]

Answer:

8.467 gm

Explanation:

The law governing this problem is "The Law of Constant Composition "

As per this law, all compounds irrespective of their origin and source have the same composition and properties in their purest form

It is a simple proportion and ratio related problem.

1.50 grams of carbon require 2.0 grams of oxygen

1.0 grams of carbon will require oxygen

= $\frac{2}{1.5}$

6.35 grams of carbon will require oxygen

$\frac{2}{1.5}\times6.35$ = 8.4666\\= 8.467

6 0

2 years ago

Between a plate and the body of a bolt, the projected area is equal to the product of the bolt _______ and the plate _______.

Elan Coil [88]

Answer:

Projected area= Diameter of the bolt* thickness.

Explanation:

Between a plate and the body of a bolt, the projected area is equal to the product of the bolt _Diameter of the bolt______ and the plate ___thickness____.

Projected area= Diameter of the bolt* thickness.

Projected area is a 2-dimensional area measurement of a 3-dimensional body by projecting its surface on an arbitrary plane

8 0

2 years ago

Topic Gravitational force amd firld strength.. help me please

I am Lyosha [343]

The gravitational force between m₁ and m₂ has magnitude

$F_{1,2} = \dfrac{Gm_1m_2}{x^2}$

while the gravitational force between m₁ and m₃ has magnitude

$F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}$

where x is measured in m.

The mass m₁ is attracted to m₂ in one direction, and attracted to m₃ in the opposite direction such that m₁ in equilibrium. So by Newton's second law, we have

$F_{1,2} - F_{1,3} = 0$

Solve for x :

$\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}$

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

$x \approx 6.74\,\mathrm m$

5 0

2 years ago

A container can hold a mass of 40 grams. If Jackson has a fake gold chain that has a density of 10 g/cm3 that occupies a space o

eimsori [14]

Use the formula M=D×V:

M=10 g/cm³ * 5 cm³ = 50 g

which is more than 40 grams, so the container cannot hold the chain.

5 0

3 years ago

In Young’s two slit experiment, the first dark fringe above the central bright fringe occurs at an angle of 0.44˚. What is the r

melomori [17]

Answer:

d / λ = 26.7

Explanation:

In Young's double slit experiment, constructive interference is described by the expression

d sin θ = m λ

In the case of destructive interference we must add half wavelength (λ/2)

d siyn θ = (m + ½) λ

Let's clear

d / λ = (m + ½) / sin θ

Let's calculate

d / λ = (2+ ½) / sin 5.4

d / λ = 5 / (2 sin 5.4)

d / λ = 26.7

8 0

2 years ago