Ask question

Ask question

All categories

2 years ago

6

Determine the identity of an unknown element X using the given table and following clues: I) Element X is a solid II) Element X

will bond ionically with element C in a 1:1 ratio. III) As a positive ion, it will have the same number of electrons as argon.

2 answers:

klasskru [66]2 years ago

6 0

The answer is calcium. I just did it and it was correct

bogdanovich [222]2 years ago

4 0

The answer here is cleary Calcium.

You might be interested in

A constant power is supplied to a rotating disc .the relationship of angular velocity of disc and number of rotations made by th

Bogdan [553]

A rotating disc supplied with constant power where the relationship of the angular velocity of the disc and the number of rotations made by the disc is governed by Newton's second law for rotation. This law is specially made for rotating bodies which is extracted from Newton's second law of motion.

6 0

3 years ago

Which of the following is true about organizing items and information during scientific investigations?

LekaFEV [45]

There are usually multiple methods for organizing items and information in a scientific investigation.

5 0

3 years ago

Read 2 more answers

Why is the picture above not an accurate representation of charges and their field lines?

pochemuha

Because field lines don’t all go in to the pole. The bottom half go outwards not inwards.

7 0

2 years ago

Read 2 more answers

A 3.00-kg object has a velocity 1 6.00 i ^ 2 2.00 j ^2 m/s. (a) what is its kinetic energy at this moment? (b) what is the net w

tatyana61 [14]

(a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:
$v= \sqrt{(6.00m/s)^2+(2.00 m/s)^2}=6.32 m/s$
And so, the kinetic energy of the object is
$K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(6.32 m/s)^2=60 J$

(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
$v= \sqrt{(8.00 m/s)^2+(4.00 m/s)^2}=8.94 m/s$
And so the new kinetic energy is
$K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(8.94 m/s)^2=120 J$

So, the work done on the object is the variation of kinetic energy of the object:
$W=\Delta K=120 J-60 J=60 J$

7 0

2 years ago

"18. Milliliter is equivalent to 1 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

Firlakuza [10]

18 mililiter is equivilent to 0.0047551

3 0

3 years ago