In classifying salts as neutral, acidic, or basic, it is important to take note of the strength of the acids and bases that they come from. A strong acid and strong base produce a neutral salt. A weak acid and strong base produce a basic salt. A strong acid and weak base produce an acidic salt. So the answers must be:
KCl = neutral (from HCl and KOH)
NH4Br = acidic (from NH4 and HBr)
K2CO3 = basic (from KOH and H2CO3)
NaCN = basic (from NaOH and HCN)
LiClO = basic (from LiOH and HClO)
Isotopes are elements with the same number of protons but different number of neutrons
Given parameters;
Mass of gold colored metal beads = 425g
Volume of water displaced by beads = 48.0cm³
Unknown;
Identity of the metal = ?
Given densities;
Gold: 19.3 g/mL
Copper: 8.86 g/mL
Bronze: 9.87 g/mL
Density is an intensive property of any substance. This implies that we can use the density of any substance to identify it.
Density can be defined as the mass per unit volume of a substance. Every substance has a unique mass per volume.
Mathematically;
Density = 
where mass is in kg or g
volume is in m³ or cm³
To find the density, we must know the mass and volume.
In this problem, the volume of the gold metal beads is the same as the volume of water displaced. This is a way to measure volume of solids.
Since the volume is given in cm³, and we are comparing with choices that have units in g/mL, we simply convert the volume in cm³ to mL
1cm³ = 1mL³
So therefore, volume of gold colored metal is 48mL
Now input the parameters given and solve for the density;
Density =
= 8.85g/mL
From the given densities, we clearly see that copper is the metal since they both of similar densities.
Answer:
1 mole is equal to 1 moles Potassium, or 39.0983 grams.
39.0983g X 4 = 156.3932
so 4 moles are in 156 grams of Potassium
Explanation:
Hope this helps
-A Helping Friend (mark brainliest pls)
Answer:
14 J of heat
Explanation:
Recall that the enthalpy of reaction is defined as the enthalpy change that occurs in a system when matter is transformed by a given chemical reaction, if all reactants and products are in their standard states. Here we have the heat of formation of six moles of the substance and we are expected to use it to obtain the the heat of formation of one mole of the compound and we can do that by simple proportion as shown below;
If 6 moles of the compound produces 84J of heat
1 mole of the substance will produce 1×84/6 = 14 J of heat