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3 years ago

How many moles of MgCl2 are there in 329 g of the compound? Your Answer:

Physics

2 answers:

Nookie1986 [14]3 years ago

6 0

Answer:3.46

Explanation:

mass=329

molar mass=71

no. of moles=329/71

=3.46

Debora [2.8K]3 years ago

5 0

Answer:3.46

Explanation:

mass=329

molar mass=71

no. of moles=329/71

=3.46

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2. What value did you calculate for the index of refraction of the acrylic block in Part 2? How does your value compare to the a

lesya [120]

The percentage error in calculated value of the refractive index of acrylic block is calculated using the formula: (true value - calculated value/true value) × 100%.

<h3>What is refractive index of a refracting medium?</h3>

The refractive index of a medium is the ratio of the sine of the angle of incidence on the medium and the sine of the angle of refraction.

Refractive index = sin i/sin r

where;

i is the angle of incidence
r is angle of refraction.

The refractive index of a medium increases with density of the medium.

To determine the percentage error in determining the refractive index of a medium, the following formula is used:

Percent error = error/true value × 100%

Therefore, the percentage error in calculated value of the refractive index of acrylic block is the ratio of the error and true value multiplied by 100%.

Learn more about refractive index at: brainly.com/question/83184

8 0

3 years ago

The ability of atoms to attract electrons is referred to as

dexar [7]

Answer: electronegativity

Explanation:

Electronegativity is defined as the property of an element to attract a shared pair of electron towards itself.

The size of an atom decreases as we move across the period because the electrons get added to the same shell and the nuclear charge keeps on increasing. Thus the electrons get more tightly held by the nucleus.

As, the size of an element decreases, the valence electrons come near to the nucleus. So, the attraction between the nucleus and the shared pair of electrons increases and thus the electronegativity increases.

8 0

4 years ago

Read 2 more answers

) Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) . What are (a)

natali 33 [55]

To develop this problem it is necessary to apply the concepts related to the Cross Product of two vectors as well as to obtain the angle through the magnitude of the angles.

The vector product between the Force and the radius allows us to obtain the torque, in this way,

$\tau = \vec{F} \times \vec{r}$

$\tau = (8i+6j)\times(-3i+4j)$

$\tau = (8*4)(i\times j)+(6*-3)(j\times i)$

$\tau = 32k +18k$

$\tau = 50 k$

Therefore the torque on the particle about the origen is 50k

PART B) To find the angle between two vectors we apply the definition of the dot product based on the vector quantities, that is,

$cos\theta = \frac{r\cdot F}{|\vec{r}|*|\vec{F}|}$

$cos\theta = \frac{(8*-3)+(4*3)}{\sqrt{(-3)^2+4^2}*\sqrt{8^2+6^2}}$

$cos\theta = -0.24$

$\theta = cos^{-1} (-0.24)$

$\theta = 103.88\°$

Therefore the angle between the ratio and the force is 103.88°

5 0

3 years ago

A step up transformer has 250 turns on its primary and 500 turns on it secondary. When the primary is connected to a 200 V and t

Zina [86]

Answer:

2000 W

Explanation:

First of all, we need to find the output voltage in the transformer, by using the transformer equation:

$\frac{V_1}{N_1}=\frac{V_2}{N_2}$

where here we have

V1 = 200 V is the voltage in the primary coil

V2 is the voltage in the secondary coil

N1 = 250 is the number of turns in the primary coil

N2 = 500 is the number of turns in the secondary coil

Solving for V2,

$V_2 = N_2 \frac{V_1}{N_1}=(500) \frac{200 V}{250}=400 V$

Now we can find the power output, which is given by

P = VI

where

V = 400 V is the output voltage

I = 5 A is the output current

Substituting,

P = (400 V)(5 A) = 2,000 W

3 0

4 years ago

If the ball leaves the projectile launcher at a speed of 2.2 m/s at an angle of 30ᴼ, and the projectile launcher is on a table a

IgorC [24]

Answer: 1.12 m

Explanation:

This situation is related to parabolic motion, hence we can use the following equations:

$y=y_{o}+V_{o}sin \theta t-\frac{g}{2}t^{2}$ (1)

$x=V_{o} cos \theta t$ (2)

Where:

$y=0 m$ is the ball final height (when it hits the ground)

$y_{o}=1.1 m$ is the ball initial height

$V_{o}=2.2 m/s$ is the initial velocity

$\theta=30\°$ is the angle at which the ball was launched

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the horizontal distance the ball travels

Rewriting (1) with the given values:

$0 m=1.1 m+(2.2 m/s)(cos 30\°)t-\frac{9.8 m/s^{2}}{2}t^{2}$ (3)

Multiplying all the eqquation by -1 and rearranging:

$4.9 m/s^{2} t^{2}-1.1 m/s t-1.1 m=0$ (4)

So, since we have a quadratic equation here (in the form of $0=at^{2}+bt+c$ , we will use the quadratic formula to find $t$ :

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ (5)

Where $a=4.9$ , $b=-1.1$ , $c=-1.1$

Substituting the known values and choosing the positive result of the equation, we have:

$t=\frac{-(-1.1)\pm\sqrt{(-1.1)^{2}-4(4.9)(-1.1)}}{2(4.9)}$

$t=0.59 s$ (6)

Now, substituting (6) in (2):

$x=(2.2 m/s)(cos 30\°)(0.59 s)$ (7)

$x=1.12 m$ (8) This is the horizontal distance at which the ball hits the ground.

3 0

4 years ago