We can collect a sample of that food and add a drop of blue litmus solution. If the color of the litmus solution changes to red, it contains an acid as one of its ingredients.
Explanation:
(a) The given data is as follows.
Pressure on top (
) = 140 bar =
(as 1 bar =
)
Temperature =
= (15 + 273) K = 288 K
Density of gas = 


= 0.4548

=
= 
Hence, pressure at the natural gas-oil interface is
.
(b) At the bottom of the tank,

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]
= 
= 309.8 bar
Hence, at the bottom of the well at
pressure is 309.8 bar.
Answer:
74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.
Explanation:
The balanced reaction is:
Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:
- Na₂CO₃: 1 mole
- Ca(NO₃)₂: 1 mole
- CaCO₃: 1 mole
- NaNO₃: 2 mole
Being the molar mass of the compounds:
- Na₂CO₃: 106 g/mole
- Ca(NO₃)₂: 164 g/mole
- CaCO₃: 100 g/mole
- NaNO₃: 85 g/mole
then by stoichiometry the following quantities of mass participate in the reaction:
- Na₂CO₃: 1 mole* 106 g/mole= 106 g
- Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
- CaCO₃: 1 mole* 100 g/mole= 100 g
- NaNO₃: 2 mole* 85 g/mole= 170 g
You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of CaCO₃?

mass of CaCO₃= 74.81 grams
<u><em>74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.</em></u>
Answer:
These particles were considered so small that they were unable to be cut
Answer:
1.23 M
Explanation:
Molarity of a substance , is the number of moles present in a liter of solution .
M = n / V
M = molarity
V = volume of solution in liter ,
n = moles of solute ,
Moles is denoted by given mass divided by the molecular mass ,
Hence ,
n = w / m
n = moles ,
w = given mass ,
m = molecular mass .
From the question ,
w = given mass of NaCl = 7.2 g
As we know , the molecular mass of NaCl = 58.5 g/mol
Moles is calculated as -
n = w / m = 7.2 g / 58.5 g/mol = 0.123 mol
Molarity is calculated as -
V = 100ml = 0.1 L (since , 1 ml = 1/1000L )
M = n / V = 0.123 mol / 0.1 L = 1.23 M