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goblinko [34]
3 years ago
12

What is the mass of the object?

Chemistry
1 answer:
galben [10]3 years ago
7 0

Answer:

37.3

263.5

Explanation:

The scale measures hundreds of units, tens of units, units, and parts of units (1 decimal place.

Scale 1

Hundreds 0 * 100 = 0

Tens: 3 * 10 = 30

Units: 7 * 1 = 7

1/10 unit = 3* 0.1 = 0.3

Total 30 + 7 + 0.3 = 37.3

Scale 2

Hundreds 2 * 100 = 200

Tens: 6 * 10 = 60

Units: 3 * 1 = 3

1/10 unit = 5* 0.1 = 0.5

Total = 200 + 60 + 3 + 0.5 = 263.5

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What is the concentration of a solution with a volume of 2.5 liters containing 600 grams of calcium phosphate?​
trapecia [35]

Answer:

1.12M

Explanation:

Given parameters:

Volume of solution  = 2.5L

Mass of Calcium phosphate  = 600g

Unknown:

Concentration  = ?

Solution:

Concentration is the number of moles of solute in a particular solution.

Now, we find the number of moles of the calcium phosphate from the given mass;

        Formula of calcium phosphate  = Ca₃PO₄

         molar mass = 3(40) + 31 + 4(16) = 215g/mol  

Number of moles of  Ca₃PO₄  = \frac{600}{215}   = 2.79moles

   Now;

  Concentration  = \frac{Number of moles }{volume }  

 Concentration  = \frac{2.79}{2.5}   = 1.12M

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3 years ago
How many blizzards are in 13m?
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7 0
3 years ago
Which phrase best summarizes how animals get water from the environment
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What are the phrases to choose from?
6 0
3 years ago
You determine the volume of your plastic bag (simulated human stomach) is 1.08 L. How many grams of NaHCO3 (s) are required to f
dsp73

Answer:

3.636 grams of sodium bicarbonate is required.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = 753.5 mmHg = 0.9914 atm

(1 atm = 760 mmHg)

V = Volume of gas = 1.08 L

n = number of moles of gas = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 24.5 °C= 297.65  K

Putting values in above equation, we get:

(0.9914 atm)\times 1.08 L=n\times (0.0821L.atm/mol.K)\times 297.65K\\\\n=0.0438 mole

Percentage recovery of carbon dioxide gas =  49.4%

Actual moles of carbon dioxide formed: 49.4% of 0.0438 mole

\frac{49.4}{100}\times  0.0438 mol=0.02164 mol

2NaHCO_3\righarrow Na_2CO_3+H_2O+CO_2

According to reaction ,1 mol is obtained from 2 moles of sodium bicarbonate.

Then 0.02164 moles f carbon dioxide will be obtained from:

\frac{2}{1}\times 0.02164 mol=0.04328 mol

Mass of 0.04328 moles pf sodium bicarbonate:

0.04328 mol × 84 g/mol = 3.636 g

3.636 grams of sodium bicarbonate is required.

5 0
4 years ago
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