I NEED HELP

Jake is in chemistry class. He makes a list of the chemicals his instructor described and the properties of each.

koban [17]

Silver: bonds with other atoms because of the weak forces of the valence electrons
FALSE - The strong forces of the valence electrons is actually the reason why silver bonds with other atoms.

Water: bonds allow for liquid state at room temperature and prevent conduction
FALSE - Water is a good conductor.

Carbon: bonds with other atoms through strong shared electrical bonds
TRUE - Carbon shares covalent bonds with other atoms.

Niobium: bonds allow for a strong conductivity found in stainless steel
FALSE - Iron and Carbon make up steel.

6 0

2 years ago

Read 2 more answers

This is “Fusion Reactions”.<br> Please answer number 8. Thank you.

Angelina_Jolie [31]

Answer:

²₁H + ³₂He —> ⁴₂He + ¹₁H

Explanation:

From the question given above,

²₁H + ³₂He —> __ + ¹₁H

Let ⁿₐX be the unknown.

Thus the equation becomes:

²₁H + ³₂He —> ⁿₐX + ¹₁H

We shall determine, n, a and X. This can be obtained as follow:

For n:

2 + 3 = n + 1

5 = n + 1

Collect like terms

n = 5 – 1

n = 4

For a:

1 + 2 = a + 1

3 = a + 1

Collect like terms

a = 3 – 1

a = 2

For X:

n = 4

a = 2

X =?

ⁿₐX => ⁴₂X => ⁴₂He

Thus, the balanced equation is

²₁H + ³₂He —> ⁴₂He + ¹₁H

8 0

2 years ago

A particular planet has a moment of inertia of 9.74 × 1037 kg ⋅ m2 and a mass of 5.98 × 1024 kg. Based on these values, what is

malfutka [58]

Answer: A) $6.38(10)^{6} m$

Explanation:

The equation for the moment of inertia $I$ of a sphere is:

$I=\frac{2}{5}mr^{2}$ (1)

Where:

$I=9.74(10)^{37}kg m^{2}$ is the moment of inertia of the planet (assumed with the shape of a sphere)

$m=5.98(10)^{24}kg$ is the mass of the planet

$r$ is the radius of the planet

Isolating $r$ from (1):

$r=\sqrt{\frac{5I}{2m}}$ (2)

Solving:

$r=\sqrt{\frac{5(9.74(10)^{37}kg m^{2})}{2(5.98(10)^{24}kg)}}$ (3)

Finally:

$r=6381149.077m \approx 6.38(10)^{6} m$

Therefore, the correct option is A.

4 0

2 years ago

A 90 kg man stands in a very strong wind moving at 17 m/s at torso height. As you know, he will need to lean in to the wind, and

victus00 [196]

Answer:

a) $t=195.948N.m$

b) $\phi=13.6 \textdegree$

Explanation:

From the question we are told that:

Density $\rho=1.225kg/m^2$

Velocity of wind $v=14m/s$

Dimension of rectangle:50 cm wide and 90 cm

Drag coefficient $\mu=2.05$

a)

Generally the equation for Force is mathematically given by

$F=\frac{1}{2}\muA\rhov^2$

$F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2$

$F=163.29$

Therefore Torque

$t=F*r*sin\theta$

$t=163.29*1.2*sin90$

$t=195.948N.m$

b)

Generally the equation for torque due to weight is mathematically given by

$t=d*Mg*sin90$

Where

$d=sin \phi$

Therefore

$t=sin \phi*Mg*sin90$

$195.948=833sin \phi$

$\phi=sin^{-1}\frac{195.948}{833}$

$\phi=13.6 \textdegree$

5 0

2 years ago

A rocket of mass m is to be launched fromplanet X, which has a mass M and a radius R. What is the minimum speed that the rocket

Sunny_sXe [5.5K]

Answer:

The minimum speed = $\sqrt{\frac{2GM}{R} }$

Explanation:

The minimum speed that the rocket must have for it to escape into space is called its escape velocity. If the speed is not attained, the gravitational pull of the planet would pull down the rocket back to its surface. Thus the launch would not be successful.

The minimum speed can be determined by;

Escape velocity = $\sqrt{\frac{2GM}{R} }$

where: G is the universal gravitational constant, M is the mass of the planet X, and R is its radius.

If the appropriate values of the variables are substituted into the expression, the value of the minimum speed required can be determined.

7 0

2 years ago