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 6.076 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 6.076 grams.

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

brainly.com/question/13073862

#SPJ1

You might be interested in

A car of mass M = 1000 kg traveling at 50.0 km/hour enters a banked turn covered with ice. The road is banked at an angle θ , an

kupik [55]

The radius of the curved road at the given condition is 54.1 m.

The given parameters:

mass of the car, m = 1000 kg
speed of the car, v = 50 km/h = 13.89 m/s
banking angle, θ = 20⁰

The normal force on the car due to banking curve is calculated as follows;

$Fcos(\theta) = mg$

The horizontal force on the car due to the banking curve is calculated as follows;

$Fsin(\theta) = \frac{mv^2}{r}$

Divide the second equation by the first;

$\frac{Fsin(\theta)}{Fcos(\theta) } = \frac{mv^2}{rmg} \\\\tan(\theta) = \frac{v^2}{rg} \\\\r = \frac{v^2}{g \times tan(\theta)} \\\\r = \frac{13.89^2}{9.8 \times tan(20)} \\\\r = 54.1 \ m$

Thus, the radius of the curved road at the given condition is 54.1 m.

Learn more about banking angle here: brainly.com/question/8169892

3 0

3 years ago

A cannonball is fired vertically upwards at 100.0 m/s a) How long will it take to return to the cannon? b) what is it's maximum

V125BC [204]

Answer:
a) 20s
b) 500m

Explanation:
Given the initial velocity = 100 m/s, acceleration = -10m/s^2 (since it is moving up, acceleration is negative), and at the maximum height, the ball is not moving so final velocity = 0 m/s.

To find time, we apply the UARM formula:

v final = (a x t) + v initial

Replacing the values gives us:

0 = (-10 x t) + 100

-100 = -10t

t = 10s

It takes 10s for the the ball to reach its max height, but it must also go down so it takes 2 trips, once going up and then another one going down, both of which take the same time to occur

So 10s going up and another 10s going down:

10x2 = 20s

b) Now that we have v final = 0, v initial = 100, a = -10, t = 10s (10s because maximum displacement means the displacement from the ground to the max height) we can easily find the displacement by applying the second formula of UARM:

Δy = (1/2)(a)(t^2) + (v initial)(t)

Replacing the values gives us:

Δy = (1/2)(-10)(10^2) + (100)(10)

= (-5)(100) + 1000

= -500 + 1000

= 500 m

Hope this helps, brainliest would be appreciated :)

7 0

3 years ago

What is the objective lens?

Monica [59]

Answer:

The objective lens is an optical tool used to focus an image.

Explanation:

The objective lens is an optical tool that collects light emitted by an object under observation and focuses the rays of light in order to form a real and magnified image They are used in optical instruments like microscopes, cameras, telescopes, etc. and are also referred as objective or object glasses.

5 0

3 years ago

Two beams of coherent light are shining on the same piece of white paper. with respect to the crests and troughs of such waves,

defon

Answer:

"where crests and troughs have their maxima at the same time"

Crests and troughs are 180 deg out of phase and when they have their maxima at the same time and place, their net contribution will be zero"

7 0

2 years ago

For a standard production car, the highest road-tested acceleration ever reported occurred in 1993, when a Ford RS200 Evolution

Ann [662]

Answer:

a = 8.06 m/s²

Explanation:

The acceleration of this car can be found using the first equation of motion:

$v_f = v_i + at\\\\a = \frac{v_f-v_i}{t}$

where,

a = acceleration = ?

vf = final speed = 26.8 m/s

vi = initial speed = 0 m/s

t = time = 3.323 s

Therefore,

$a = \frac{26.8\ m/s-0\ m/s}{3.323\ s}$

a = 8.06 m/s²

3 0

3 years ago