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3 years ago

I need help finding number 2

Chemistry

1 answer:

Amiraneli [1.4K]3 years ago

7 0

you did the 1st one wrong!

1 mole of Zn gives 1 mole of ZnCl2

and 1.5 moles of Zn gives 1.5 moles of ZnCl2.

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Which of the following metals would be the best for the new frying pan?

Rudik [331]

Answer:

Copper

Explanation:

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3 years ago

What would be the mass of a 33.5dm3 sample of O2 at STP?

liraira [26]

Answer:

• One mole of oxygen is equivalent to 16 grams.

→ But at STP, 22.4 dm³ are occupied by 1 mole.

${ \tt{22.4 \: {dm}^{3} \: \dashrightarrow \: 16 \: grams}} \\ { \tt{33.5 \: {dm}^{3} \: \dashrightarrow \: ( \frac{33.5 \times 16}{22.4} ) \: grams}} \\ \\ \dashrightarrow \: { \boxed{ \tt{23.94 \: g \approx \: 24 \: grams}}}$

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3 years ago

The following diagram of a pendulum shows the path it takes as it swings back and forth.

lbvjy [14]

I believe it should be B

4 0

3 years ago

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Identify the formula for an alkene.

ioda

Answer:

CnH2n-2

im pretty sure thats the answer

5 0

3 years ago

A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.62 g of water at 52.3 oC in an insulated container. clear = 0.128

alisha [4.7K]

Answer: The final temperature of both the weight and the water at thermal equilibrium is $50.26^{o}C$ .

Explanation:

The given data is as follows.

mass = 7.62 g, $T_{2} = 10.8^{o}C$

Let us assume that T be the final temperature. Therefore, heat lost by water is calculated as follows.

q = $mC \times \Delta T$

= $7.62 g \times 4.184 J/^{o}C \times (52.3 - T)$

Now, heat gained by lead will be calculated as follows.

q = $mC \times \text{Temperature change of lead}$

= $2.04 \times 0.128 \times (T - 11.0)$

According to the given situation,

Heat lost = Heat gained

$7.62 g \times 4.184 J/^{o}C \times (52.3 - T)$ = $2.04 \times 0.128 \times (T - 11.0)$

T = $50.26^{o}C$

Thus, we can conclude that the final temperature of both the weight and the water at thermal equilibrium is $50.26^{o}C$ .

7 0

3 years ago