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4 years ago

Explain the connections between reactants, products, and limiting reactant.

1 answer:

6 0

Reactants is a starting substance in a chemical reaction; (reactants appear on the left-hand side of a chemical equation).
Products is a substance produced in a chemical reaction; (products appear on the right-hand side of a chemical equation).
Limiting reactant is the reactant that has the smallest stoichiometric amount in a reactant mixture and consequently limits the amount of product in a chemical reaction.

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A shell of mass m and speed v explodes into two identical fragments. If the shell was moving horizontally (the positive x direct

-Dominant- [34]

Answer:

The velocity of the other fragment immediately following the explosion is v .

Explanation:

Given :

Mass of original shell , m .

Velocity of shell , + v .

Now , the particle explodes into two half parts , i.e $\dfrac{m}{2}$ .

Since , no eternal force is applied in the particle .

Therefore , its momentum will be conserved .

So , Final momentum = Initial momentum

$mv=\dfrac{mv}{2}+\dfrac{mu}{2}\\\\u=v$

The velocity of the other fragment immediately following the explosion is v .

4 0

3 years ago

You move a 4 N object 10 meters. Find work

Trava [24]

Answer: 40 J

Explanation: Work is equal to the product of force and distance.

W = Fd

= 4N ( 10 m)

= 40 J

8 0

4 years ago

Tarzan, who weighs 849 N, swings from a cliff at the end of a 18.0 m vine that hangs from a high tree limb and initially makes a

loris [4]

Answer:

Part a)

$T = 342.5 \hat i + 675\hat j$

Part b)

$F_{net} = 342.5\hat i - 174\hat j$

Part c)

$F = 384.2 N$

Part d)

$\theta = 333 degree$

Part e)

$a = 4.4 m/s^2$

Part f)

$\theta = 333 degree$

Explanation:

Part a)

Magnitude of tension force is given as

$T = 757 N$ at 26.9 degree with vertical

$T = 757 sin26.9 \hat i + 757 cos26.9 \hat j$

$T = 342.5 \hat i + 675\hat j$

Part b)

Net force on Tarzen is given as

$F_{net} = T + F_g$

$F_{net} = 342.5 \hat i + 675\hat j - 849 \hat j$

$F_{net} = 342.5\hat i - 174\hat j$

Part c)

magnitude of the force is given as

$F = \sqrt{F_x^2 + F_y^2}$

$F = \sqrt{342.5^2 + 174^2}$

$F = 384.2 N$

Part d)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-174}{342.5}$

$\theta = 333 degree$

Part e)

Magnitude of the acceleration

$a = \frac{F}{m}$

$m = \frac{849}{9.81} = 86.5 kg$

tex]a = \frac{384.2}{86.5}[/tex]

$a = 4.4 m/s^2$

Part f)

Direction of acceleration is same as the direction of the force

$\theta = 333 degree$

6 0

3 years ago

IF U ANSWER I WILL PUT U AS BRAINIEST ANSWER AND GIVE U A THANK U

Bingel [31]

<span>
At the Earth's surface, warm air expands and rises, creating
what is known as an area of low pressure.

Cold air is dense and sinks to the surface to create what is
known as an area of high pressure.</span>

8 0

4 years ago

Read 2 more answers

Please help. Having a hard time figuring out

Goryan [66]

Yeah that’s is correct

5 0

3 years ago