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just olya [345]
3 years ago
8

Use the reaction below to balance the nuclear reaction equation.

Chemistry
2 answers:
Temka [501]3 years ago
5 0

Answer:

neon-22

Explanation:

Temka [501]3 years ago
4 0

Answer:

neon-22

Explanation:

i got it right..have a great day.Jesus loves you<3

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What is the formula used to calculate the heat required to warm or cool one phase of matter
Ronch [10]

Answer:

SPECIFIC HEAT CAPACITY

Explanation:

Q = MC DELTA T

q = heat

c = specific heat

T = temperature ( final - initial )

4 0
4 years ago
CLIC 3. The blank is the central part of an atom.​
lorasvet [3.4K]

Answer:

The nucleus (center) of the atom contains the protons (positively charged) and the neutrons (no charge). The outermost regions of the atom are called electron shells and contain the electrons (negatively charged).

6 0
3 years ago
What is Titan? Why are scientists interested in it?
Pavlova-9 [17]
Titan is one of Saturn's moon and they are interested in it because it has an atmosphere like Earth.
5 0
4 years ago
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In the following reaction, which species is reduced? Au(s) + 3NO3-(aq) + 6H+(aq) → Au3+(aq) + NO(g) + 3H2O (l) H+ N+5 O2- H2O Au
weqwewe [10]

Answer:

NO3-

Explanation:

Given the reaction equation;

Au(s) + 3NO3-(aq) + 6H+(aq)→Au3+(aq) + 3NO2(g) + 3H2O (l).

We can consider the oxidation states of species on the left and right hand sides of the reaction equation;

Au is in zero oxidation state on the left hand side and an oxidation state of +3 on the righthand side.

NO3- is in oxidation state of +5 on the righthand side and NO2 is in + 4 oxidation state.

H+ is in + 1 oxidation state on both the left and right hand sides of the reaction equation.

Since reduction has to do with a decrease in oxidation number, it follows that NO3- was reduced in the reaction.

6 0
3 years ago
A hot lump of 115.7 g of an unknown substance initially at 168.3°C is placed in 25.0 mL of water initially at 25.0°C and allowed
marshall27 [118]

The substance has the specific heat capacity of steel and is therefore probably steel.  

Let the specific heat of the unknown lump of substance be c.

Energy Exchange = Specific Heat ⨯ Mass ⨯ Temperature Change

Energy the Hot Lump Lost = Energy the Cold Water Gained

Water has a specific heat of 4.2 \; \text{J} \cdot \text{g} ^{-1}\cdot \text{K}^{-1} and a density of 1 \; \text{g} \cdot \text{ml}^{-1}. 25.0 milliliters of water thus has a mass of 25.0 grams.

115.7 \times (168.3 - 76.5) \; c = 4.2 \times 25.0 \times (76.5 - 25.0) \\c = 0.51 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}

Steel has a specific heat of approximately 0.51 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}. This substance is thus <em>probably </em>steel.


4 0
4 years ago
Read 2 more answers
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