All categories

3 years ago

How many moles of calcium (Ca) atoms are there in 77.4 g of Ca?

2 answers:

8 0

Lar mass of Ca = 40.08 grams/mole 77.4 g Ca * ( 1 mole Ca/ 40.08 ... n = m / M 1mol Caweights 40 gmol-1 n = 77,4 / 40 = 1.93 mol.

iren [92.7K]3 years ago

3 0

Answer: 1.94 moles

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

Given mass = 77.4 g

Molar mass = 40 g/mol

Putting in the values:

$\text{Number of moles}=\frac{77.4g}{40g/mol}=1.94$

Thus 1.94 moles of calcium atoms are there in 77.4 g of Ca.

You might be interested in

Which subatomic particle has negligible mass and travels around outside the nucleus?

nikdorinn [45]

Electrons travel around the nucleus

6 0

3 years ago

Use the ruler to determine the length of the line measured to the correct degree of precision. 13.2 cm 13.25 cm 130 cm 132 mm

lapo4ka [179]

Your answer would be 13.25cm

Hope that helps!!!

3 0

3 years ago

Read 2 more answers

Suppose you begin with an unknown volume of 8.61 m h2so4 and add enough water to make 5.00*102 ml of a 1.75 m h2so4 solution. wh

olga55 [171]

1.02x10^2 ml Since molarity is defined as moles per liter, the product of the molarity and volume will remain constant as mole solvent is added. So let's set up an equality to express this m0*v0 = m1*v1 where m0, v0 = molarity and volume of original solution m1, m1 = molarity and volume of final solution. Solve for v0, then substitute the known values and calculate: m0*v0 = m1*v1 v0 = (1.75 M * 500 ml)/8.61 M v0 = (1.75 M * 500 ml)/8.61 M V0 = 101.6260163 Rounding to 3 significant figures gives 102 ml. So the original volume of the 8.61 M H2SO4 solution was 102 ml or 1.02x10^2 ml.

3 0

3 years ago

Most minerals on earth's surface contain what

muminat

The answer is silicates!!! Hope this helps!!!

8 0

3 years ago

Patient added a 17-g measured dose of polyethylene glycol 3350 (MIRALAX) to 180 mL of water to use as a laxative. If the volume

Sergio [31]

Answer: Apparent density of ethylene glycol 3350 = 17g/15.6mL= 1.09g/mL

Specific gravity of mixture = 1.01

Explanation:

The specific gravity of an object or substance is the ratio of the density of that object or substance to that of a reference substance usually water.

Specific gravity = density of substance/density of water

The density of a substance is the ratio of the mass of that substance to is volume.

Density = Mass/volume

Density of water 1.00 g/mL,

Apparent density of ethylene glycol 3350 = mass/volume

volume of ethylene glycol = volume of mixture - volume of water

volume of ethylene glycol = 195.6 - 180) = 15.6mL

Apparent density of ethylene glycol 3350 = 17g/15.6mL= 1.09g/mL

mass of water = density of water * volume of water

mass of water = 1.00g/mL * 180mL = 180 g

Mass of solution = mass of substance + mass of water = (17 + 180)g = 197g

Density of mixture = mass of mixture / volume of mixture

Density of mixture = 197 g / 195.6 mL = 1.01 g/mL

specific gravity of mixture = density of mixture / density of water

specific gravity of mixture = 1.01g/mL / 1.00g/mL

specific gravity of mixture = 1.01

8 0

3 years ago