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sweet-ann [11.9K]
3 years ago
14

Predict the Solubility of the following substances in water.

Chemistry
1 answer:
anyanavicka [17]3 years ago
5 0
<span>a. NaNO3: soluble
b. AgBr: insoluble
c. NH4OH: soluble
d. Ag2CO3: insoluble
e. NH4Br: soluble
f. BaSO4: insoluble
g. Pb(OH)2: soluble
h. PbCO3: insoluble</span>
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Oxygen and hydrogen are both elements that are found as gases at room temperature. When oxygen combines with hydrogen, they prod
Tom [10]

Answer:

A

Explanation:

has properties that are different from the original substances.

5 0
2 years ago
a particular application calls for N2 g with a density of 1.80 g/L at 32 degrees C what must be the pressure of the n2 g in mill
baherus [9]

Answer:

1223.38 mmHg

Explanation:

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

Also,  

Moles = mass (m) / Molar mass (M)

Density (d)  = Mass (m) / Volume (V)

So, the ideal gas equation can be written as:

PM=dRT

Given that:-

d = 1.80 g/L

Temperature = 32 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (32 + 273.15) K = 305.15 K

Molar mass of nitrogen gas = 28 g/mol

Applying the equation as:

P × 28 g/mol  = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K

⇒P = 1223.38 mmHg

<u>1223.38 mmHg must be the pressure of the nitrogen gas.</u>

5 0
3 years ago
How to do limiting reagents for chemistry?
gavmur [86]
Balance the chemical equation for the chemical reaction.
Convert the given information into moles.
Use stoichiometry for each individual reactant to find the mass of product produced.
The reactant that produces a lesser amount of product is the limiting reagent.
The reactant that produces a larger amount of product is the excess reagent.
To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given.
4 0
3 years ago
Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.76
aliina [53]

The given question is incomplete, the complete question is:

Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.763g of oxygen. For a given mass of oxygen, what is the lowest whole-number mass ratio of lead in the two compounds that combines with a given mass of oxygen?

Answer:

The lowest whole-number mass ratio in the two compounds is 1:2.

Explanation:

There is a need to find the mole ratio between lead and oxygen atoms in order to find the whole-number mass ratio of lead in the two compounds. In the first compound, the given mass of lead is 2.98 grams, the molar mass of lead is 207.2 gram per mole.  

The no. of moles can be determined by using the formula,  

moles = mass/molecular mass

moles = 2.98 g/207.2 g/mol

= 0.0144 moles

The mass of oxygen in the compound I is 0.461 grams, the molecular mass of oxygen is 16 gram per mol.  

moles = 0.461 g /16 g/mol

= 0.0288 moles

The ratio between the lead and oxygen in the compound I is 0.0144/0.0288 = 1:2

On the other hand, in the compound II, the mass of lead given is 9.89 grams, therefore, the moles of lead in compound II is,  

moles = 9.89 g / 207.2 g/mol

= 0.0477 moles

The mass of oxygen given in compound II is 0.763 grams, the moles of oxygen present in the compound II is,  

moles = 0.763 g / 16 g

= 0.0477 moles

The ratio between the lead and oxygen in the compound II is, 0.0477 moles lead /0.0477 moles oxygen = 1:1

Hence, of the two compounds, the lowest ratio is found in the compound I, that is, 1:2.  

4 0
2 years ago
A scientist needs to collect a 0.050 mole sample of helium, but needs to know how large his helium container should be. What vol
Mumz [18]

We can use the ideal gas equation:

PV = nRT

P = 202.6kPa = 202600 Pa (You have to multiply by 1000)

n = 0.050 mole

R = 0.082 atm*l/(K*mol)

T = 400K

We will have to convert from Pa to atm or viceversa.

101325 Pa________1 atm

202600 Pa________x = 2.00 atm

2atm*V = 0.050 mole*0.082 atm*l/(K*mol)* 400K

V = 0.050 mole*0.082 atm*l/(K*mol)* 400K/2atm = 0.82 liters = 820 mililiters



8 0
3 years ago
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