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Molodets [167]
2 years ago
11

1. I’m in the 2nd column, 4th row, and I’m a metal. Who am I? ________________ 2. I’m a very lonely nonmetal. Who am I? ________

____ 3. I’m the only metal who is a liquid at room temperature. Who am I? ____________ 4. I’m named after the person who created the 1st Periodic Table. Who am I? ___________ 5. I have 92 protons. Who am I? _____________ 6. I’m the only nonmetal who is a liquid at room temperature. Who am I? ___________ 7. I’m named after a very famous scientist. Who am I? ___________ 8. I have 46 electrons. Who am I? ____________ 9. My atomic mass is 183.84. Who am I? _____________ 10. My chemical symbol is Ag. Who am I? ________________ 11. I’m the only metalloid in period 3. Who am I? ___________ 12. I’m the only element that is solid and a nonmetal in group 14. Who am I? _____________ 13. I have 5 neutrons. Who am I? ____________ 14. I’m the only gas at room temperature that is in group 16. Who am I? ___________ 15. I have 68 protons. Who am I? __________ 16. What element has the chemical symbol of Ir? ______________ 17. Which element is in group 7 and has 30 neutrons. Who am I? ___________ 18. I’m the only metal in group 15. Who am I? ____________ 19. I have 88 electrons. Who am I? ___________ 20. I’m the only gas at room temperature and in period 5. Who am I? ____________ 21. My symbol is Am. Who am I? ______________ 22. I’m the only nonmetal in period 6. Who am I? ____________ 23. My atomic number is 69.723. Who am I? _________________ 24. I have 159 neutrons. Who am I? ________________ 25. I’m the only metalloid in group 17. Who am I? ______________ 26. I have 50 electrons. Who am I? __________________ 27. I’m in the 1st group and the 4th period. Who am I? ________________ 28. I’m a metalloid whose symbol is Sb. Who am I? ______________ ©JFlowers2017 Name: ______________________________ Date: ___________Class: ________ Periodic Table Scavenger Hunt Directions: You will use the Periodic Table to answer the questions. 1. I’m in the 17th column, a nonmetal, & a solid at room temperature. Who am I? ________________ 2. I have 79 electrons. Who am I? ____________ 3. I’m the only gas in period 6. Who am I? ____________ 4. My atomic mass is 257. Who am I? ___________ 5. My chemical symbol is Hs. Who am I? _____________ 6. I have 114 neutrons. Who am I? ___________ 7. I’m in the 18th group and 2 nd period. Who am I? ___________ 8. I have 67 protons. Who am I? ____________ 9. I’m a nonmetal who is solid at room temperature & has 2 letters for my symbol. Who am I? _________ 10. I’m in the 1 st group & 7 th period. Who am I? ________________ 11. I’m the only metalloid in group 13. Who am I? ___________ 12. I have 97 electrons. Who am I? _____________ 13. I am the only gas in column 15. Who am I? ____________ 14. My name is similar to Mickey Mouse’s best friend. Who am I? ___________ 15. I’m in group 11 & period 4. Who am I? __________ 16. I have 62 protons. Who am I? ______________ 17. My name fits really well with doctors because they try to do this. Who am I? ___________ 18. My name reminds me of where we all live. Who am I? ____________ 19. I’m the only nonmetal in period 2. Who am I? ___________ 20. My atomic number is 87.62. Who am I? ____________ 21. My symbol is Mt. Who am I? ______________ 22. I’m in group 17 & the only metalloid. Who am I? ____________ 23. I have 71 electrons. Who am I? _________________ 24. My symbol is Pd. Who am I? ________________ 25. I’m Dorothy’s friend who needed a heart. Who am I? ______________ 26. I have 41 protons. Who am I? __________________ 27. I have 125 neutrons. Who am I? ________________ 28. My name comes from the 8th planet. Who am I? ______________
Physics
1 answer:
Ghella [55]2 years ago
6 0

Answer:

Explanation:

I'm in 17th column , a nometal, and a solid at room temperature. What am i

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A soccer field is viewed from above, while a ball is kicked eastward with an initial speed of 10.0 m/s. The ball experiences a c
satela [25.4K]

Answer:

the ball travelled approximately 60 m towards north before stopping

Explanation:

 Given the data in the question;

First course : a_{x} = 0.75 m/s², d_{x} = 20 m, u_{x} = 10 m/s

now, form the third equation of motion;

v² = u² + 2as

we substitute

v_{x}² = (10)² + (2 × 0.75 × 20)

v_{x}² = 100 + 30

v_{x}² = 130

v_{x} = √130

v_{x} = 11.4 m/s

for the Second Course:

u_{y} =  11.4 m/s,  a_{y} = -1.15 m/s²,  v_{y} = 0

Also, form the third equation of motion;

v² = u² + 2as

we substitute

0² = (11.4)² + (2 × (-1.15) ×  d_{y} )

0 = 129.96 - 2.3d_{y}

2.3d_{y}  = 129.96

d_{y} = 129.96 / 2.3

d_{y} = 56.5 m

so;

|d| = √( d_{x}² + d_{y}² )

we substitute

|d| = √( (20)² + (56.5)² )

|d| = √( 400 + 3192.25 )

|d| = √( 3592.25 )

|d| = 59.9 m ≈ 60 m

Therefore, the ball travelled approximately 60 m towards north before stopping

7 0
2 years ago
A football is kicked from the ground with a velocity of 38m/s at an angle of 40 degrees and eventually lands at the same height.
Anastasy [175]

Initially, the velocity vector is \langle 38cos(40^{\circ}),38sin(40^{\circ}) \rangle=\langle 29.110, 24.426 \rangle. At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by 4.9(0.2)^2, so the velocity is \langle 29.110, 24.426-0.196 \rangle = \langle 29.110, 24.23 \rangle.

Converting back to direction and magnitude, we get \langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle

4 0
3 years ago
A spy in a speed boat is being chased down a river by government officials in a faster craft. Just as the officials’ boat pulls
babunello [35]

Answer:

They are 7.4m apart.

Explanation:

Here we have a parabolic motion problem. we need the time taken to land so:

Y=Yo+Vo*t+\frac{1}{2}*a*t^2

considerating only the movement on Y axis:

0=4.6-(9.81)*t^2\\t=0.68s

Because we have a contant velocity motion on X axis:

xs=vs*t\\xs=17m/s*(0.68s)\\xs=11.6m

and

xg=vg*t\\xg=28m/s*(0.68s)\\xg=19m

the distance between them is given by:

d=|xg-xs|\\d=7.4m

7 0
3 years ago
A car approaches you at a constant speed, sounding its horn, and you hear a frequency of 76 Hz. After the car goes by, you hear
Talja [164]

Answer:

70.07 Hz

Explanation:

Since the sound is moving away from the observer then

f_o = f_s\frac {(v+vs)}{v} and f_o = f_s\frac {(v-vs)}{v} when moving towards observer

With f_o of 76 then taking speed in air as 343 m/s we have

76 = f_s\times\frac {(343-vs)}{343}

f_s=\frac {343\times 76}{343-v_s}

Similarly, with f_o of 65 we have

65 = f_s\times\frac {(343+vs)}{343}\\f_s=\frac {343\times 65}{343+v_s}

Now

f_s=\frac {343\times 65}{343+v_s}=\frac {343\times 76}{343-v_s}

v_s=27.76 m/s

Substituting the above into  any of the first two equations then we obtain

f_s=70.07 Hz

4 0
3 years ago
Give three factors which are responsible for the vanishing forest ​
slava [35]

Answer:

1. Huge wildfires

2. Deforestation

3. Reduced amount of aforestation, etc

5 0
3 years ago
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