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Lorico [155]
3 years ago
6

The density of a certain type of steel is 8.1 g/cm3. What is the mass of a 124 cm3 chunk of this steel? Show

Chemistry
1 answer:
posledela3 years ago
8 0

Mass of the steel is 1004.4 g

  • density of a substance can be regarded as the ratio of the mass of that substance to that of it's volume.

  • Mass of substance can be regarded as the quantitative measure of inertia, it can be simply explained as a fundamental property of all matter

<em> We can express this mathematically as ;</em>

  • Density=  \frac{ mass  }{volume  }

Given :

Density of this steel= 8.1g/cm3

Mass=  124 cm^{3}

If we make Mass subject of formula,

Mass= ( density *volume)

Then let us substitute the values into the expression

Mass= (124*8.1)

= 1004.4 g

Therefore, mass of the steel is 1004.4 g

Learn more at: brainly.com/question/9196460?referrer=searchResults

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tangare [24]

Answer:

D

Explanation:

depending on the solid the shape will not change.

3 0
3 years ago
What volume (mL) of 0.135 M NaOH is required to neutralize 13.7 mL of 0.129 M HCl? a: 0.24 b: 13.1 c: 0.076 d: 6.55 e: 14.3
Len [333]

Answer:

The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).

Explanation:

The reaction between an acid and a base is called neutralization, forming a salt and water.

Salt is an ionic compound made up of an anion (positively charged ion) from the base and a cation (negatively charged ion) from the acid.

When an acid is neutralized, the amount of base added must equal the amount of acid initially present. This base quantity is said to be the equivalent quantity. In other words, at the equivalence point the stoichiometry of the reaction is exactly fulfilled (there are no limiting or excess reagents), therefore the numbers of moles of both will be in stoichiometric relationship. So:

V acid *M acid = V base *M base

where V represents the volume of solution and M the molar concentration of said solution.

In this case:

  • V acid= 13.7 mL= 0.0137 L (being 1,000 mL= 1 L)
  • M acid= 0.129 M
  • V base= ?
  • M base= 0.135 M

Replacing:

0.0137 L* 0.129 M= V base* 0.135 M

Solving:

V base=\frac{0.0137 L*0.129 M}{0.135 M}

V base=0.0131 L = 13.1 mL

<u><em> The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).</em></u>

4 0
3 years ago
Hi! I produce proteins inside the cell. What am I?​
Arturiano [62]

Answer:

ribosome is the cell

Explanation:

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6 0
3 years ago
Read 2 more answers
150 ml of 0.1 m naoh is added to 200 ml of 0.1 m formic acid, and water is added to give a final volume of 1 l. what is the ph o
N76 [4]

Number of moles of NaOH = V(NaOH) * M(NaOH)= 0.150 L * 0.1 moles/L = 0.015 moles

Number of moles of formic acid, HCOOH = V(HCOOH) * M(HCOOH) = 0.200 L * 0.1 moles/L = 0.020 moles

Here, the limiting reagent is NaOH

The reaction is represented as:

HCOOH + NaOH ↔HCOONa + H2O

Moles of HCOONa formed = Moles of the limiting reagent, NaOH = 0.015 moles

Moles of HCOOH remaining = 0.020-0.015 = 0.005 moles

Total final volume is given as 1 L

Therefore: [HCOOH] = 0.005 moles/1 L = 0.005 M

[HCOONa] = 0.015/1 = 0.015 M

pKa of HCOOH = 3.74

As per Henderson-Hasselbalch equation

pH = pka + log[HCOONa]/[HCOOH] = 3.74+log[0.015/0.005] = 4.22

Therefore, pH of the final solution = 4.22


                       


3 0
3 years ago
What is the mass of Al₂O3 that will contain 10 kg of aluminium?​
Oksanka [162]

Answer:

1 mole of Al2O3 = 102 grams

1 mole of Al2 = 54 grams

102 grams of Al2O3 contains = 54 gram of Al2

10kg of Al2O3 contains = (54/102)*10000g Al2

= 5294.11 g Al2 or 5.29411 kg

5 0
2 years ago
Read 2 more answers
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