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Illusion [34]
3 years ago
12

Sodium nitrate is used in fertilizers, explosives, and glass manufacturing. Give the formula for this compound

Chemistry
1 answer:
krek1111 [17]3 years ago
7 0

Explanation:

The molecular formula for Sodium Nitrate is NaNO3. The SI base unit for amount of substance is the mole. 1 grams Sodium Nitrate is equal to 0.011765443644878 mole.

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What elements does lithium resemble (in terms of electrons) when it loses 1 electron?
tigry1 [53]

Answer:

Li atoms readily give up one electron to form positively charged, Li+ ions. These ions have the same stable electron configuration as the noble gas helium. All Group 1 atoms can lose one electron to form positively charged ions.

Explanation:

5 0
3 years ago
In an exothermic reaction, the bonding energy of the product is:
ahrayia [7]
I think the correct answer from the choices listed above is option A. <span>In an exothermic reaction, the bonding energy of the product is </span><span>less than the reactant because it is only at this condition that the energy is released by the reaction.</span>
3 0
3 years ago
Read 2 more answers
In order to study hydrogen halide decomposition, a researcher fills an evacuated 1.11 L flask with 0.822 mol of HI gas and allow
Elza [17]

Answer:

Kc = 168.0749

Explanation:

  •           2HI(g)     ↔    H2(g) + I2(g)

initial mol:   0.822               0          0

equil. mol: 2(0.822 - x)         x           x

∴ [ HI ]eq = 0.055 mol/L = 2(0.822 - x) / (1.11 L )

⇒ 1.644 - 2x = 0.055 * 1.11

⇒ 1.644 = 2x + 0.06105

⇒ 2x = 1.583

⇒ x = 0.7915 mol equilibrium

⇒ [ H2 ] eq = 0.7915mol / 1.11L = 0.7130 M = [ I2 ] eq

⇒ Kc = ([ H2 ] * [ I2 ]) / [ HI ]²

⇒ Kc = ( 0.7130² ) / ( 0.055² )

⇒ Kc = 168.0749

 

4 0
3 years ago
A 50.0 mL sample of an aqueous H2SO4 solution is titrated with a 0.375 M NaOH solution. The equivalence point is reached with 62
vesna_86 [32]

Answer: The concentration of H_2SO_4 is 0.234 M

Explanation:

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity H_2SO_4 = 2

M_1 = molarity of H_2SO_4 solution = ?

V_1 = volume of  H_2SO_4 solution = 50.0 ml

n_2 = acidity of NaOH = 1

M_1 = molarity of NaOH solution = 0.375 M

V_1 = volume of  NaOH solution = 62.5 ml

Putting in the values we get:

2\times M_1\times 50.0=1\times 0.375\times 62.5

M_1=0.234M

Therefore concentration of H_2SO_4 is 0.234 M

6 0
3 years ago
Simon has collected three samples from the coral reef where he observes marine life. He must determine whether each one is a pur
-Dominant- [34]

Answer:

SAMPLE A - pure substance.

SAMPLE B - homogeneous mixture.

SAMPLE C - heterogeneous mixture.

Explanation:

3 0
3 years ago
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