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3 years ago

Which is an IUPAC name for a covalent compound

Physics

2 answers:

leonid [27]3 years ago

4 0

Answer:

carbon dioxide on 2020 edge..

Explanation:

just got a 100% on the test

Elena L [17]3 years ago

3 0

In naming covalent compound (binary) based in IUPAC naming, we have 4 rules to be followed:

1. The first element of the formula will use the normal name of the given element. for example: CO2 ( Carbon Dioxide), Carbon is the element name of the first element of the formula.

2. The second element is named as if they are treated like an anion but put in mind that these are no ions in a covalent compound but we put -ide on the second element as if it is an anion.

3. Prefixes are used to indicate the number of atom of the elements in the compound. for example: mono- 1 atom, di- 2atoms, tri- 3 atoms and etc

4. Prefix "mono"is never used in naming the first element. For example: Carbon dioxide, there should be no monocarbon dioxide.

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phase, color, and ductility are all examples of what type of property? chemical changeable physical external

svet-max [94.6K]

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3 years ago

What are Sir Issac Newton's three laws of motion?

Likurg_2 [28]

The first law is that every object stay at rest or stay in uniform motion in a straight line until it is forced to change its state by the action of an external force. This law is called law of inertia.

The second law is that the acceleration of an object is dependent upon two variables. the net force acting upon the object and the mass of the object. F= ma or force is equal to mass times acceleration. This law is known as the law of force and acceleration.

The third law is that for every action there is an equal and opposite reaction. every interaction there is a pair of forces acting on the two interacting objects. the size of forces on the first object equals the size of the force on the second object.

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can you please make this the brainliest answer it would really help . Thanks

4 0

3 years ago

Read 2 more answers

A 460 g , 6.0-cm-diameter can is filled with uniform, dense food. It rolls across the floor at 1.1 m/s . Part A What is the can'

Reika [66]

Answer:

the can's kinetic energy is 0.42 J

Explanation:

given information:

Mass, m = 460 g = 0.46 kg

diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m

velocity, v = 1.1 m/s

the kinetic energy of the can is the total of kinetic energy of the translation and rotational.

KE = $\frac{1}{2}$ I ω^2 + $\frac{1}{2} mv^{2}$

where

I = $\frac{1}{2} mr^{2}$ and ω = $\frac{v}{r}$

thus,

KE = $\frac{1}{2}$ $\frac{1}{2} mr^{2}$ ( $\frac{v}{r}$ )^2 + $\frac{1}{2} mv^{2}$

= $\frac{1}{2}$ $\frac{1}{2} mr^{2}$ $\frac{v^{2} }{r^{2}}$ + $\frac{1}{2} mv^{2}$

= $\frac{1}{4} mv^{2}$ + $\frac{1}{2} mv^{2}$

= $\frac{3}{4} mv^{2}$

= $\frac{3}{4} (0.46) (1.1)^{2}$

= 0.42 J

8 0

3 years ago

A cart loaded with bricks has a total mass of 9.13 kg and is pulled at constant speed by a rope. The rope is inclined at 24.7 ◦

blagie [28]

Answer:

W = 0.63 KJ

Explanation:

Work (W) is defined as the point product of force (F) by the distance (d)the body travels due to this force.

W= F*d *cosα Formula (1)

F : force (N)

d : displacement (m)

α : angle between force and displacement

Newton's second law:

∑F = m*a Formula (2)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the cart on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the cart

W: Weight of the cart : In vertical direction

FN : Normal force : perpendicular to the floor

f : Friction force: parallel to the floor

T : tension Force, inclined at θ=24.7° above the horizontal

Calculated of the W

W= m*g

W= 9.13 kg* 9.8 m/s² = 89.47 N

x-y components o the tension force (T)

Tx = Tcosθ = T*cos 24.7° (N)

Ty = Tsin θ = T*sin 24.7° (N)

Calculated of the FN

We apply the formula (2)

∑Fy = m*ay ay = 0

FN +Ty- W = 0

FN = W-Ty

FN = 89.47-T*sin 24.7°

Calculated of the friction force (f)

f = μk*FN

f =(0.597)*( 89.47-T*sin 24.7° )

f= 53.41-0.249T

Calculated of the tension force of the rope (f)

We apply the formula (2) :

∑Fx = m*ax , ax= 0 ,because the speed of the cart is constant

Tx - f = 0

T*cos 24.7°-( 53.41 - 0.249T )= 0

T*cos 24.7° + 0.249T = 53.41

(1.1575)T = 53.41

T= (53.41) / (1.1575)

T= 46.14 N

Work done on the cart by the rope

We apply the formula (1)

W=T*d *cosα

W= (46.14 N)*(15.1 m) *(cos24.7)

W = 632.97 (N*m) = 632.97 (J)

W = 0.63 KJ

6 0

4 years ago

Can any kind soul help me please

Alex

Answer:

I) 420000J

ii)

Explanation:

(I) so you can use the formula for quantity of heat then substitute the values given

formula-Q=mc∆9

3 0

3 years ago