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Arturiano [62]
2 years ago
12

A sample of a certain material has a mass of 2.03 × 10–3 g. Calculate the volume of the sample, given that the density is 9.133

× 10 –1 g/cm3 . Use the four-step method to solve the problem.
Chemistry
2 answers:
Vaselesa [24]2 years ago
8 0

Answer:

Volume of the material given is 0.0022cm^{3}

Explanation:

The relation between density, mass and volume is:

density=\frac{mass}{volume}

Given:

Density = 9.133X10^{-1}\frac{g}{cm^{3}}

mass = 2.03X10^{-3}g

Therefore the volume will be:

volume=\frac{mass}{density}=\frac{2.03X10^{-3}}{9.133X10^{-3}}=0.0022cm^{3}

Volume of the material given is 0.0022cm^{3}

liq [111]2 years ago
3 0
2.22x10^-3 would be the answer to the question 
please mark as brainliest answer 
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A container holds 15.0 g of phosphorous gas at a pressure of 2.0 atm and a temperature of 20.0 Celsius. What is the density of t
erik [133]
<span>Density is a value for mass, such as kg, divided by a value for volume, such as m3. Density is a physical property of a substance that represents the mass of that substance per unit volume. We calculate as follows:

PV = nRT
PV = mRT/ Molar mass
m/V = P(molar mass)/RT
Density = P(molar mass)/RT 
Density = 2.0 ( 30.97 ) / 0.08206 ( 20 + 273.15) = 2.57 g/L <----First option</span>
5 0
2 years ago
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14) The central iodine atom in the ICl4- ion has __________ nonbonded electron pairs and __________ bonded electron pairs in its
masha68 [24]

Answer:

Two non bonded electron pairs and four bonded electron pairs

Explanation:

An image of the compound as obtained from chemlibretext is attached to this answer.

The ion ICl4- ion, is an AX4E2 ion. This implies that there are four bond pairs and two lone pairs of electrons. As expected, the shape of the ion is square planar since the lone pairs are found above and below the plane of the square. This is clear from the image attached.

7 0
2 years ago
Which equation shows the proper setup to find the volume of a 0.250 M CaCl2 solution that contains 39.5 g of CaCl2?
Jlenok [28]
Molarity = number of mole of substance(n) / volume of solution (V).

n(CaCl2) = mass (CaCl2)/M(CaCl2)

M(CaCl2) = 40+2*35.5 = 111 g/mol 

n(CaCl2) =39.5 g CaCl2*1 mol/111g

0.250 M = 39.5 g CaCl2*1 mol/111g*volume of solution (V).

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4 0
2 years ago
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A 1.87 L aqueous solution of KOH contains 155 g of KOH . The solution has a density of 1.29 g/mL . Calculate the molarity ( M ),
lbvjy [14]

Answer:

[KOH] = 1.47 M

[KOH] = 1.22 m

KOH = 6.86 % m/m

Explanation:

Let's analyse the data

1.87 L is the volume of solution

Density is 1.29 g/mL → Solution density

155 g of KOH → Mass of solute

Moles of solute is (mass / molar mass) = 2.76 moles.

Molarity is mol/L → 2.76 mol / 1.87 L = 1.47 M

Let's determine, the mass of solvent.

Molality is mol of solute / 1kg of solvent

We can use density to find out the mass of solution

Mass of solution - Mass of solute = Mass of solvent

Density = Mass / volume

1.29 g/mL = Mass / 1870 mL

Notice, we had to convert L to mL, cause the units of density.

1.29 g/mL . 1870 mL = Mass → 2412.3 g

2412.3 g - 155 g = 2257.3 g of solvent

Let's convert the mass of solvent to kg

2257.3 g / 1000 = 2.25kg

2.76 mol / 2.25kg = 1.22 m (molality)

% percent by mass = mass of solute in 100g of solution.

(155 g / 2257.3 g) . 100g = 6.86 % m/m

5 0
2 years ago
When balanced, which equation would have the coefficient 3 in front of any of the reactants? Zn + HCl ZnCl2 + H2 H2SO4 + B(OH)3
Gwar [14]

The correct answer is B. H2SO4 + B(OH)3  B2(SO4)3 + H2O

Hope this helps!

6 0
2 years ago
Read 2 more answers
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