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nikklg [1K]
3 years ago
15

How many bonds and lone pairs are in the molecule SIH4?

Chemistry
1 answer:
lesantik [10]3 years ago
6 0

Answer:

there are no valence electrons left over, so the molecule has four bond pairs and no lone pairs.

Explanation:

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PLEASE HELP!!! I'll give brainliest answer
ELEN [110]

Answer:

1. MG

2. N

3. F

4. NA

5. O

6. K

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2 years ago
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The graph indicates the running route for Tobias.
Musya8 [376]

Answer: b

Explanation:

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Which of the following is in intensive property a. mass b. magnetism c shape D. volume
lisov135 [29]

Answer:

b. Magnetism (sorry im very late)

Explanation:

Intensive properties do not depend on size, no matter what it doesn't. For example, magnetism, density, melting and boiling points, and color. All of those support intensive property.

3 0
3 years ago
Identify the type of reaction shown by this chemical equation:<br>2AI + 6HCI - 2AlCl3 + 3H2​
ArbitrLikvidat [17]

Answer:

Single displacement reaction

Explanation:

2AI + 6HCI —> 2AlCl3 + 3H2

From the above reaction, we see clearly that Al displaces H from HCl. This is clearly a single displacement reaction.

5 0
2 years ago
Cesium-137 has a half life of 30.0 years. If initially there are 8.0 kg of cesium-137 present in a sample, how many kg will rema
viva [34]

Mass after 60 years:

2 kg of cesium-137 will remain after 60.0 years.

What is Half-life?

The duration needed for a quantity to decrease to half of its initial value is known as the half-life.

Given:

N0 = is the initial amount of cesium-137 = 8.0 Kg

N = is the amount remaining after a time =  Unknown

t =  Duration of cesium-137 decay in 8.0 Kg = 60 years

t1/2 = half-life of cesium-137 = 30.0 years

Formula = N/No = (1/2)^t/t1/2

After putting the value:

N/8= (1/2)^60/30

N/8= (1/2)^2

N/8= 1/4

4N= 8

N= 8/4

N= 2 Kg

As a result, 2 Kg of cesium-137 will be present after 60 years.

Learn more about the half-life here,

brainly.com/question/14018709

#SPJ4

8 0
2 years ago
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