Ask question

Ask question

All categories

dimulka [17.4K]

3 years ago

8

In the following reaction, which substance is the precipitate? (NH4)2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2NH4NO3(aq) (NH4)2SO4 Ba(

NO3)2 BaSO4 NH4NO3

2 answers:

Arte-miy333 [17]3 years ago

5 0

Answer:

BaSo4

Explanation:

Mariana [72]3 years ago

4 0

I think it's Barium sulfate, the soild and percipitate

You might be interested in

The laws of thermodynamics state that, in a heat engine, _____.

olchik [2.2K]

<span>There are four laws of thermodynamic which define and characterize the thermodynamic system at thermal equilibrium.
The laws of thermodynamics state that, in a heat engine, </span>all the heat energy from a source cannot be converted to mechanical energy.

4 0

3 years ago

Read 2 more answers

a car that is moving in a straight line has an acceleration of 8.0 m/s/s for 4.0 seconds. which one of the following is true of

Nuetrik [128]

Answer:

32 ms

Explanation:

v=32ms.

Explanation:

I will assume that you mean that the acceleration is 8.0 ms2, as 8.0ms is a value for velocity, not acceleration.

Here, we use the formula v=u+at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is time. Let's substitute values, then:

v=u+at,

v=0+8.0⋅4.0

v=32,

v=32ms.

Hope it Helps! :D .

7 0

2 years ago

Your answer is CORRECT. Let v⃗ be the vector with initial point (8,1) and terminal point (6,4). Express v⃗ as a linear combinati

Phoenix [80]

Answer:

1) Option E is correct.

vector v = (-2î + 3ĵ)

2) Option C is correct.

The vertical component of vector v = 2

3) Option B is correct.

The vector sum of u and v = (4î - 9ĵ)

4) Option B is correct.

5u - 4v = (-30î - -23ĵ)

5) Option F is correct.

Magnitude of v = √32 units = 4√2 units = 5.66 units

6) Option F is correct.

Unit vector in the same direction as v is

v = (î + 3ĵ)/√10 = [(1/√10), (3/√10)]

Explanation:

1) vector v has initial point (8,1) and terminal point (6,4)

Write vector v as a linear combination of the standard unit vector.

v represented in standard form is given as

Vector v = (final position vector) - (initial position vector) = (6î + 4ĵ) - (8î + ĵ) = (-2î + 3ĵ)

2) v be the vector with initial point (−5,1) and terminal point (5,3). Find the vertical component of this vector.

v represented in standard form is given as

Vector v = (final position vector) - (initial position vector) = (5î + 3ĵ) - (-5î + ĵ)

v = (10î + 2ĵ)

The vertical component is the ĵ-component and it is equal to 2

3) Find the sum of the vectors u =6i −4j and v⃗ =−2i −5j

Vector sum is done on a per component basis

Sum = u + v = (6î - 4ĵ) + (-2î - 5ĵ) = (4î - 9ĵ)

4) Given vectors u⃗ =⟨−2,−3⟩ and v⃗ =⟨5,2⟩; find 5u⃗ −4v⃗

u = (-2î - 3ĵ)

v = (5î + 2ĵ)

5u - 4v = 5(-2î - 3ĵ) - 4(5î + 2ĵ)

= (-10î - 15ĵ) - (20î + 8ĵ)

= (-30î - 23ĵ)

5) Find the magnitude of the vector v = (4i−4j)

Magnitude of a vector is given as

/v/ = √[vₓ² + vᵧ²]

where vₓ and vᵧ are x and y components of the velocity.

/v/ = √[(4²) + (-4)²] = 4√2 units = 5.66 units

6) Given the vector v =⟨1,3⟩; find a unit vector in the same direction as v

Unit vector in the direction of a vector = (vector)/(magnitude of vector)

Vector v = (î + 3ĵ)

Magnitude of vector v = √[1² + 3²] = √10

Unit vector in the same direction as v = (î + 3ĵ)/√10

Hope this Helps!!!

8 0

3 years ago

Please help <br>problems 2a.,2b.,3a.,and 3b.

Darina [25.2K]

Answer:

2a) x = 32 [mil/h]; 2b) t = 0.5[h]; 3a) t = 2.5 [h]; 3b) x = 185[mil]

Explanation:

2a)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

$v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 32 [\frac{mil}{h}] \\t=time = 1 [h]\\x=v*t\\x=32[\frac{mil}{h} ]*1[h]\\x=32[mil}$

2b)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

$v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{420}{840}\\ t=0.5[h]$

3a)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

$v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{35}{14}\\ t=2.5[h]$

3b)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

$v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 74 [\frac{mil}{h}] \\t=time = 2.5 [h]\\x=v*t\\x=74[\frac{mil}{h} ]*2.5[h]\\x=185[mil}$

8 0

3 years ago

A stuntwoman is going to attempt a jump across a canyon that is 77 m wide. The ramp on the far side of the canyon is 25 m lower

liq [111]

initial speed of the stuntman is given as

$v = 28 m/s$

angle of inclination is given as

$\theta = 15 degree$

now the components of the velocity is given as

$v_x = 28 cos15 = 27.04 m/s$

$v_y = 28 sin15 = 7.25 m/s$

here it is given that the ramp on the far side of the canyon is 25 m lower than the ramp from which she will leave.

So the displacement in vertical direction is given as

$\delta y = -25 m$

$\delta y = v_y * t + \frac{1}{2} at^2$

$-25 = 7.25 * t - \frac{1}{2}*9.8* t^2$

by solving above equation we have

$t = 3.12 s$

Now in the above interval of time the horizontal distance moved by it is given by

$d_x = v_x * t$

$d_x = 27.04 * 3.12 = 84.4 m$

since the canyon width is 77 m which is less than the horizontal distance covered by the stuntman so here we can say that stuntman will cross the canyon.

5 0

3 years ago