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miv72 [106K]
3 years ago
8

What is 1 Oz equal to in grams?

Chemistry
2 answers:
lakkis [162]3 years ago
4 0
1 Oz is 28.3495 grams

hope this helps!
pentagon [3]3 years ago
3 0

Conversion factor for ounces and grams.

<em>1 ounce is equal to approximately 28.4 grams. </em>

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A gas has experienced a small increase in volume but has maintained the same pressure and number of moles. According to the idea
krek1111 [17]

Answer:

If a gas has experienced a small increase in volume but has maintained the same pressure and number of moles, the temperature of the gas will DROP.

Explanation:

According to Boyle’s law of ideal gases, volume and temperature of a gas is inversely related, as long as the pressure is kept constant;

P₁V₁/T₁ = P₂V₂/T₂

Therefore, if the volume of the gas increases, the temperature will definitely decrease due to the inverse relationship. The gas will get cooler.  

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3 years ago
Which of the following is the correct name for the compound HF?
alexira [117]
Name for the compound HF is Hydrogen fluoride.
Hope this helps!
6 0
3 years ago
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Given 9.0 moles of O2, how many grams of H2O can be produced ?
Anastasy [175]

Answer:

If we have one mole of water, then we know that it will have a mass of 2 grams (for 2 moles of H atoms) + 16 grams (for one mole O atom) = 18 grams.

Explanation:

5 0
3 years ago
Why was the international system of units adopted?
marissa [1.9K]
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3 years ago
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the solubility product Ag3PO4 is: Ksp = 2.8 x 10^-18. What is the solubility of Ag3PO4 in water, in moles per liter?
guapka [62]

Answer : The solubility of Ag_3PO_4 in water is, 1.8\times 10^{-5}mol/L

Explanation :

The solubility equilibrium reaction will be:

Ag_3PO_4\rightleftharpoons 3Ag^++PO_4^{3-}

Let the molar solubility be 's'.

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^{+}]^3[PO_4^{3-}]

K_{sp}=(3s)^3\times (s)

K_{sp}=27s^4

Given:

K_{sp} = 2.8\times 10^{-18}

Now put all the given values in the above expression, we get:

K_{sp}=27s^4

2.8\times 10^{-18}=27s^4

s=1.8\times 10^{-5}M=1.8\times 10^{-5}mol/L

Therefore, the solubility of Ag_3PO_4 in water is, 1.8\times 10^{-5}mol/L

3 0
3 years ago
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