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Ugo [173]
3 years ago
10

Study this equation carefully. What classification should this reaction have? Cu + 2AgNO3 Cu(NO3)2 + 2Ag synthesis decomposition

single replacement double displacement neutralization
Physics
2 answers:
mario62 [17]3 years ago
8 0

Answer;

-Single replacement

Explanation;

The reaction;

Cu(s) + 2AgNO3= Cu(NO3)2 + 2Ag , is an example of a single replacement reaction.

-It is a type of chemical reaction where an element reacts with a compound and takes the place of another element in that compound.

In the above reaction copper(Cu) takes the place of silver(Ag) in the salt AgNO3 to form Copper (ii) nitrate (CuNO3) and silver metal (Ag).

- Another example of a single replacement reaction occurs when potassium (K) reacts with water (H2O). A colorless solid compound named potassium hydroxide (KOH) forms, and hydrogen gas (H2) is set free.


zepelin [54]3 years ago
5 0

Answer:

single replacement

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Which of the following is a heterogeneous mixture? A. salt B. dye in water C. sugar water D. a garden salad
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3 years ago
Choose all facts that increase the orbital velocity of a vessel around planet B. Bigger mass of planet B smaller mass of planet
telo118 [61]

Answer:

- Bigger mass of planet B  

- orbiting closer to planet B

Explanation:

The orbital velocity of the vessel around the planet can be found by equalizing the force of gravity between the vessel and the planet and the centripetal force:

G\frac{mM}{r^2}=m\frac{v^2}{r}

where

G is the gravitational constant

m is the mass of the vessel

M is the mass of the planet

r is the distance between the vessel and the centre of the planet

v is the orbital velocity of the vessel

Re-arranging the formula, we find an expression for v:

v=\sqrt{\frac{GM}{r}}

We see that:

- the bigger the mass of the planet, M, the bigger the velocity

- the bigger the distance between the vessel and the planet, r, the smaller the velocity

So, the correct choices that increase the orbital velocity are:

- Bigger mass of planet B  

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6 0
3 years ago
a car is moving 8.80 m/s when it begins to accelerate at 2.45 m/s^2. how much time does it take to trav 138m. please help me (':
Ulleksa [173]

Answer:

7.6 s

Explanation:

Considering kinematics formula for final velocity as

v^{2}=u^{2}+2as

Where v and u are final and initial velocities, a is acceleration and s is distance moved.

Making v the subject then

v=\sqrt{u^{2}+2as}

Substituting 8.8 m/s for u, 138 m for s and 2.45 m/s2 for a then

v=\sqrt{8.8^{2}+2*2.45*138}\\v=27.45 m/s

Also, v=u+at and making t the subject of the formula

t=\frac {v-u}{a}

Substituting 27.45 m/s for v, 8.8 m/s for u and 2.45 m/s for a then

t=\frac {27.45-8.8}{2.45}=7.6122448979591\approx 7.6s

Therefore, it needs 7.6 seconds to travel

7 0
3 years ago
A charge q produces an electric field of strength 2E at a distance of d away. Determine the electric field strength at a distanc
Alja [10]

The electric field strength of a point charge is inversely proportional to the square of the distance from the charge ... a lot like gravity.

If the magnitude of the field is (2E) at the distance 'd', then at the distance '2d', it'll be (2E)/(2²).  That's (2E)/4 = 0.5E .

3 0
3 years ago
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