All categories

2 years ago

Type the correct answer in the box. Express your answer to three significant figures.

Physics

1 answer:

Marrrta [24]2 years ago

5 0

The mass of calcium phosphate the reaction can produce is <u>6.076</u> grams.

Mass is a numerical measure of inertia, which is a basic feature of all matter. It is, in effect, a body of matter's resistance to a change in speed or position caused by the application of a force.

In the International System of Units (SI), the kilogram is the unit of mass.

Given;

Mass of calcium nitrate (Ca(NO₃)₂) = 96.1 g

Mass of calcium phosphate=?

The chemical equation is found as;

3Ca(NO₃)₂ + 2Na₃PO₄ → 6NaNO₃ + Ca₃(PO4)₂

3 moles of Ca(NO₃)₂ produces 1 mole of Ca₃(PO4)molar mass of calcium phosphate is 164 g/mol

Moles are found as the ratio of the mass of the compound and its molar mass. Moles of Ca(NO₃)₂ is n.

n = 96.1/164 = 0.5859 moles

Moles of Ca3(PO4)2 produced ;

N=0.0589 ×(1/3) = 0.0196 moles

The molar mass of calcium phosphate is 310 g/mol and the mass of calcium phosphate produced will be;

M=0.0196×310

M = 6.076 g

Hence,the mass of calcium phosphate the reaction can produce is <u>6.076</u> grams.

To learn more about the mass, refer to the link;

brainly.com/question/13073862

#SPJ1

You might be interested in

What type of wave travels through matter only??

Olenka [21]

A mechanical wave can only travel through matter.

7 0

3 years ago

Read 2 more answers

If you wanted to find a sample of fermium, which has an atomic number of 100, where would you look?

Umnica [9.8K]

Answer:

Explanation:

Fermium is a synthetic element with the symbol Fm and atomic number 100. It is an actinide and the heaviest element that can be formed by neutron bombardment of lighter elements, and hence the last element that can be prepared in macroscopic quantities, although pure fermium metal has not yet been prepared.[3] A total of 19 isotopes are known, with 257Fm being the longest-lived with a half-life of 100.5 days.

It was discovered in the debris of the first hydrogen bomb explosion in 1952, and named after Enrico Fermi, one of the pioneers of nuclear physics. Its chemistry is typical for the late actinides, with a preponderance of the +3 oxidation state but also an accessible +2 oxidation state. Owing to the small amounts of produced fermium and all of its isotopes having relatively short half-lives, there are currently no uses for it outside basic scientific research.

5 0

2 years ago

Read 2 more answers

An electron is placed on a line connecting two fixed point charges of equal charge but the opposite sign. The distance between t

viktelen [127]

Answer:

a) F_net = 6.48 10⁻¹⁸ ( $\frac{1}{x^2} + \frac{1}{(0.300-x)^2}$ ), b) x = 0.15 m

Explanation:

a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

F = $k \frac{q_2q_2}{r^2}$

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron r₁ = x

right side -electron r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

F_net = k q Q ( $\frac{1}{r_1^2}+ \frac{1}{r_2^2}$ )

F _net = kqQ ( $\frac{1}{x^2} + \frac{1}{(d-x)^2}$ )

let's substitute the values

F_net = 9 10⁹ 1.6 10⁻¹⁹ 4.50 10⁻⁹ ( $\frac{1}{x^2} + \frac{1}{(0.30-x)^2}$ )

F_net = 6.48 10⁻¹⁸ ( $\frac{1}{x^2} + \frac{1}{(0.300-x)^2}$ )

now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values of the distance (x) and the net force are given

x (m) F (N)

0.05 27.0 10-16

0.10 8.10 10-16

0.15 5.76 10-16

0.20 8.10 10-16

0.25 27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m

4 0

3 years ago

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

where

I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$

$\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)$

$\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m$

Therefore the torque exerted on it is $2.45*10^{-3}N\cdot m$

3 0

3 years ago

Two resistors, A and B, are connected in series to a 6.0 V battery. A voltmeter connected across resistor A measures a potential

mestny [16]

Answer:

Resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

Explanation:

When the two resistors are in series, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B.

Given that V₁ + V₂ = 6.0 V and V₁ = 4.0 V,

V₂ = 6.0 V - V₁ = 6.0 V - 4.0 V = 2.0 V

Also, let the current in series be I.

So, V₁ = IR₁ and V₂ = IR₂

I = V₁/R₁ and I = V₂/R₂

equating both expressions, we have

V₁/R₁ = V₂/R₂

4.0 V/R₁ = 2.0 V/R₂

dividing through by 2.0 V, we have

2/R₁ = 1/R₂

taking the reciprocal, we have

R₂ = R₁/2

R₁ = 2R₂

From the parallel connection, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B. Since it is parallel, V₁ = V₂ = V = 6.0 V

Also, V₂ = I₂R₂ where I₂ = current in resistor B = 2.0 A and R₂ = resistance of resistor B

So, R₂ = V₂/I₂

= 6.0 V/2.0 A

= 3.0 Ω

R₁ = 2R₂

= 2(3.0 Ω)

= 6.0 Ω

So, resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

6 0

3 years ago