6th grade science. Help me please !! :))

determine the maximum amount of NaN03 that was produced during the experiment. Explain how you determined the amount

Angelina_Jolie [31]

Answer:

9 moles of NaNO3 is obtained

Explanation:

The balanced chemical reaction equation for the reaction is;

Al(NO3)3 + 3NaCl-------> 3NaNO3 + AlCl3

Now, we have to determine the limiting reactant. The limiting reactant yields the least amount of NaNO3.

1 mole of Al(NO3)3 yields 3 moles of NaNO3

4 moles of Al(NO3)3 yields 4 * 3/1 = 12 moles of NaNO3

Also,

3 moles of NaCl yields 3 moles of NaNO3

9 moles of NaCl yields 9 * 3/3 = 9 moles of NaNO3

Hence, NaCl is the limiting reactant and 9 moles of NaNO3 is obtained.

7 0

3 years ago

Lithium and nitrogen react in a combination reaction to produce lithium nitride: 6Li (s) + N2 (g) → 2Li3N (s) In a particular ex

seraphim [82]

Answer: 6.71 g

Explanation: $6Li(s)+N_2(g)\rightarrow 2Li_3N$

${\text{no of moles}}=\frac{\text{Given mass}}{\text{Molar mass}}$

${\text {moles of lithium}}=\frac{4g}{6.914g/mol}=0.578moles$

$\text{moles of nitrogen}=\frac{4g}{28g/mol}=0.143moles$

Limiting reagent is the reagent which limits the formation of product. Excess reagent is one which is in excess and thus remains unreacted.

Thus lithium is the limiting reagent and nitrogen is the excess reagent.

As can be seen from the balanced chemical equation, 6 moles of lithium reacts with 1 mole of nitrogen to give 2 moles of lithium nitride.

Thus 0.578 moles of lithium react with 0.096 moles of nitrogen.

6 moles of lithium give = 2 moles of lithium nitride

Thus 0.578 moles of lithium give= $\frac{2}{6}\times {0.578}=0.19moles$ of lithium nitride.

Mass of lithium nitride $Li_3N={\text {no of moles}}\times {\text {Molecular mass}}$

Mass of lithium nitride $Li_3N$ = ${0.192moles}\times {34.83g/mol}=6.71g$

8 0

3 years ago

A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 4

Stella [2.4K]

Answer : The final equilibrium temperature of the water and iron is, 537.12 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of iron = 560 J/(kg.K)

$c_1$ = specific heat of water = 4186 J/(kg.K)

$m_1$ = mass of iron = 825 g

$m_2$ = mass of water = 40 g

$T_f$ = final temperature of water and iron = ?

$T_1$ = initial temperature of iron = $352^oC=273+352=625K$

$T_2$ = initial temperature of water = $20^oC=273+20=293K$

Now put all the given values in the above formula, we get:

$(825\times 10^{-3}kg)\times 560J/(kg.K)\times (T_f-625K)=-(40\times 10^{-3}kg)\times 4186J/(kg.K)\times (T_f-293K)$

$T_f=537.12K$

Therefore, the final equilibrium temperature of the water and iron is, 537.12 K

8 0

3 years ago

John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bottle,

enot [183]

One can solve the problem by using the law of conservation of momentum. The total momentum prior to the collision must be equivalent to the total momentum after the collision, so we have:

m1v1 + m2v2 = m1v1 + m2v2

Here, m1 is 0.4 Kg that is the mass of the ball, u1 is 18 m/s that is the initial velocity of the ball, m2 is 0.2 Kg that is the mass of the bottle, and u2 is 0 that is the initial velocity of the bottle.

v1 is the final velocity of the ball, which is to be determined, and v2 is 25 m/s that is the final velocity of the bottle.

Substituting and rearranging the equation, one can find the final velocity of the ball:

v1 = m1u1 - m2v2 / m1 = (0.4 kg) (18 m/s) - (0.2 Kg) (25 m/s) / 0.4 Kg = 5.5 m/s.

5 0

3 years ago

Read 2 more answers

Definition of Solid State of Matter

Oksanka [162]

Solid state of matter
- the atoms are tightly packed in orderly form, they slightly vibrate but are stuck in a fixed position. Solids cannot be compressed and they maintain their shape

5 0

3 years ago