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Fittoniya [83]
3 years ago
8

When an ionic compound is formed, is energy gained or lost?

Chemistry
1 answer:
Mrrafil [7]3 years ago
6 0

Answer:

<u><em>neither</em></u>

Explanation:

it is <u><em>neither</em></u> because the <u><em>law of conservation of mass states that matter can not be created nor destroyed only rearranged!!!</em></u>

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H2S (g) + O2(g) -&gt; SO2+ H2O, how many grams of O2 are needed to make 25.0 g of SO2 (Hint: balance the equation first)
vekshin1

Answer:

Hello

The answer is 18,7(b)

Explanation:we don't know the value of o² so at first we should put (x)instead of gr of o²and then write 1molO²/32grO²×2mol SO²/3mol O²×64gr SO²/1molSO²=25 gr SO².and then just find the value of (x).

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3 years ago
PLS HELP THIS IS TIMED
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3 years ago
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How many milliliters of 0.150 M NaOH are required to neutralize 85.0 mL of 0.300 M H2SO4 ? The balanced neutralization reaction
nekit [7.7K]

Answer : The volume of NaOH required to neutralize is, 340 mL

Explanation :

To calculate the volume of base (NaOH), we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.300M\\V_1=85.0mL\\n_2=1\\M_2=0.150M\\V_2=?

Putting values in above equation, we get:

2\times 0.300M\times 85.0mL=1\times 0.150M\times V_2\\\\V_2=340mL

Hence, the volume of NaOH required to neutralize is, 340 mL

4 0
4 years ago
During dry weather, the water level in a small pond drops because of
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Water levels tend to drop in dry weather because of evaporation. The hot temperatures will cause the water to achieve its gaseous state and go up into the atmosphere.
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3 years ago
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a) \: n = \frac{N}{NA} \\ 1.5 = \frac{x}{6.02 \times \: {10}^{23} } \\ x = 1.5 \times 6.02 \times {10}^{23} \\ x = 9.03 \times {10}^{23} \: atoms \: of \: Na

b) \: n = \frac{N}{NA} \\ 0.25 = \frac{x}{6.02 \times \: {10}^{23} } \\ x = 0.25 \times 6.02 \times {10}^{23} \\ x = 1.505 \times {10}^{23} \: atoms \: of \: Si \:
c) \: n = \frac{N}{NA} \\ 6.755 = \frac{x}{6.02 \times \: {10}^{23} } \\ x = 6.755 \times 6.02 \times {10}^{23} \\ x = 40,6651 \times {10}^{23} \\ x = 4.07 \times {10}^{24} \: atoms \: of \: Pb \:
5 0
3 years ago
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