Answer:
6.76 mol/L
Explanation:
A chemist prepares a solution 0.607 kg of calcium bromide by measuring out of calcium bromide into a 450. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's calcium bromide solution. Be sure your answer has the correct number of significant digits.
Step 1: Given data
Mass of calcium bromide (solute): 0.607 kg (607 g)Volume of solution (V): 450. mLStep 2: Calculate the moles of solute
The molar mass of calcium bromide is 199.89 g/mol.
607 g × 1 mol/199.89 g = 3.04 mol
Step 3: Convert "V" to liters
We will use the conversion factor 1 L = 1000 mL.
450. mL × 1 L/1000 mL = 0.450 L
Step 4: Calculate the concentration of calcium bromide in mol/L
[CaBr₂] = 3.04 mol/0.450 L = 6.76 mol/L
HELPPPP PLZ
A student pushes a 40-N block across the floor for a distance of 10 m.
How much work was done to move the block?
400 Pascals
400 Newtons
400 Joules
400 Watts
400 Joules
Work done is measured in Joules (J) remember that
Answer:400 joules
wut do u need
Explanation:
Does warm air rise or fall?
rise and cold air
it doesn't fall cause I already fall inlove with levi
If 1000g of water from Luboc River contains 0.0075g of Ca2+ ion, what is the concentration in ppm by mass of Ca2+ present in Luboc River?
7.5 ppm
Further explanationGiven
1000 g of water
0.0075g of Ca²⁺ ion
Required
the concentration in ppm by mass of Ca²⁺
Solution
ppm = part per million
solvent = water ⇒ ppm = 1 mg/L(water density is 1 kg / L) or mg/kg
Convert g to mg of Ca²⁺ ion :
0.0075 g = 7.5 x 10⁻³ g = 7.5 mg
Convert g to kg of water :
1000 g = 1 kg water
So the concentration of Ca²⁺ ion :
= 7.5 mg / 1 kg
= 7.5 ppm
If two reactant molecules collide with each other what two reasons might they not combine ?
How many moles of O2 are in 125 grams of CO2?
(this is a gram to mole conversion) HELP ME PLEASE!!!
Answer:
20 mole of co2
Explanation:
I hope this helps
What is the theoretical yield of SO3 produced by 8.96 g of S?
Answer: Theoretical yield of [tex]SO_3[/tex] produced by 8.96 g of S is 33.6 g
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} S=\frac{8.96g}{32g/mol}=0.28moles[/tex]
The balanced chemical equation is:
[tex]2S+3O_2\rightarrow 2SO_3[/tex]
According to stoichiometry :
2 moles of [tex]S[/tex] produce = 3 moles of [tex]SO_3[/tex]
Thus 0.28 moles of [tex]S[/tex] will produce=[tex]\frac{3}{2}\times 0.28=0.42moles[/tex] of [tex]SO_3[/tex]
Mass of [tex]SO_3=moles\times {\text {Molar mass}}=0.42moles\times 80g/mol=33.6g[/tex]
Thus theoretical yield of [tex]SO_3[/tex] produced by 8.96 g of S is 33.6 g
There are
molecules of carbon dioxide (CO2) in 102.5 grams.
Answer:
1.403x10²⁴ molecules
Explanation:
In order to calculate how many molecules of CO₂ are there in 102.5 g of the compound, we first convert grams to moles using its molar mass:
102.5 g ÷ 44 g/mol = 2.330 mol CO₂Now we convert moles into molecules using Avogadro's number:
2.330 mol * 6.023x10²³ molecules/mol = 1.403x10²⁴ moleculesConvert 5.802 g/cm^3 to Kg/L
Answer:
5.80200 Kg / L
Explanation:
If one iron nail weighs 2.0 grams, what would be the mass of 6.022 x 10^23 nails?
Answer:
[tex]1.2x10^{24}g[/tex]
Explanation:
Hello!
In this case, it is possible to treat this problem by using a proportional factor which indicates one iron nail equals 2.0 grams:
[tex]\frac{2.00g}{1nail}[/tex]
Now, for an amount of 6.022x10²³ nails, the corresponding mass will be:
[tex]6.022x10^{23}nail*\frac{2.00g}{1nail} \\\\1.2x10^{24}g[/tex]
Best regards!
HRISTINA HERRERA: Attempt 1
Question 9 (2.5 points)
3
Which of the following sets of data are consistent with the law of conservation of
matter?
6
7.5 g of hydrogen gas reacts with 50.0 g oxygen gas to form 57.5 g of water.
9
50 g gasoline reacts with 243 g oxygen to form 206 g of carbon dioxide and 97 g
water
12
17.7 g nitrogen react with 34.7 g oxygen to form 62.4 g nitrogen dioxide.
all of these
15
none of these
Answer:
7.5 g of hydrogen gas reacts with 50.0 g oxygen gas to form 57.5 g of water.
Explanation:
Here we have the check if the mass of the reactants is equal to the mass of the products.
Reactants
[tex]7.5+50=57.5\ \text{g}[/tex]
Products
[tex]57.5\ \text{g}[/tex]
The data is consistent with the law of conservation of matter.
Reactants
[tex]50+243=293\ \text{g}[/tex]
Products
[tex]206+97=303\ \text{g}[/tex]
The data is not consistent with the law of conservation of matter.
Reactant
[tex]17.7+34.7=52.4\ \text{g}[/tex]
Products
[tex]62.4\ \text{g}[/tex]
The data is not consistent with the law of conservation of matter.
Only the first data is consistent with the law of conservation of matter.