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krok68 [10]
3 years ago
11

What direction are the groups (families) in the periodic table?

Chemistry
2 answers:
Soloha48 [4]3 years ago
6 0

Answer:

left

Explanation:

Troyanec [42]3 years ago
6 0
The answer to the question is left
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From the combinations of substances listed below, which would most likely be miscible in
polet [3.4K]

Answer:

C

Explanation:

polar has unequal sharing of electrons that has the lone pairs which has the electronegativity difference. can be mixed with water.

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3 years ago
Substance A and B are the same mass and at the same initial temperature. When the same amount of heat is added to each, the fina
Bingel [31]
That both of them are warm....at a certain temputurature causing one to fell,cozy
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3 years ago
An electron:
Korvikt [17]
It has a negative charge emits energy when it moves to a lower energy orbit from an excited state and it has the same mass as a proton
4 0
4 years ago
A baby weighs 7 pounds, 4 ounces at birth and 7 pounds, I ounce at discharge. What percent of weight did the baby lose
Aloiza [94]

Answer:

2.6%

Explanation:

As, 1 ounce (oz) =  0.0625 pounds (lb)

Therefore, weight of baby at discharge = 7 lb,1 oz = 7+0.0625 lb = 7.0625 lb

Since, 1 oz =  0.0625 lb

⇒ 4 oz = 4×0.0625 = 0.25 lb

Therefore, weight of baby at birth = 7 lb,4 oz = 7+0.25 lb = 7.25 lb

The <u>amount of weight lost</u> is equal to the difference of weight of the baby at birth and discharge.

Therefore, <u>weight lost</u> = 7.25 lb - 7.0625 lb = <u>0.1875 lb</u>

Now, the <u>percentage of weight lost</u> by the baby is given by the amount of weight lost divided by the weight of the baby at birth.

Therefore, <u>the percentage of weight los</u>t = weight lost ÷ weight at birth = 0.1875 lb ÷ 7.25 lb × 100 = <u>2.6% </u>

6 0
4 years ago
What is the entropy change in the environment when 5.0 MJ of energy is transferred thermally from a reservoir at 1000 K to one a
Leni [432]

Answer:

The entropy change in the environment is 3.62x10²⁶.

Explanation:

The entropy change can be calculated using the following equation:

\Delta S = \frac{Q}{k_{B}}(\frac{1}{T_{f}} - \frac{1}{T_{i}})

Where:

Q: is the energy transferred = 5.0 MJ

k_{B}: is the Boltzmann constant = 1.38x10⁻²³ J/K  

T_{i}: is the initial temperature = 1000 K

T_{f}: is the final temperature = 500 K

Hence, the entropy change is:

\Delta S = \frac{5.0 \cdot 10^{6} J}{1.38 \cdot 10^{-23} J/K}(\frac{1}{500 K} - \frac{1}{1000 K}) = 3.62 \cdot 10^{26}                                    

Therefore, the entropy change in the environment is 3.62x10²⁶.

I hope it helps you!          

7 0
3 years ago
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