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Vedmedyk [2.9K]
3 years ago
10

Unicellular organisms are made of multiple cells. O True O False

Chemistry
2 answers:
11111nata11111 [884]3 years ago
7 0

Answer:false

Explanation:unicellular organisms mean organisms made up of single cell eg;bacteria,some protists and multicellular organisms mean organisms made up of multiple cells eg;animals, plants

butalik [34]3 years ago
5 0
Its True. Because Uni means more than one. Hope this helped :)
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Analysis of a gaseous chlorofluorocarbon, CClxFy, shows that it contains 11.79% C and 69.57% Cl. In another experiment, you find
uranmaximum [27]

Answer:

The molecular formula = C_2Cl_{4}F_2

Explanation:

Moles =\frac {Given\ mass}{Molar\ mass}

% of C = 11.79

Molar mass of C = 12.0107 g/mol

<u>% moles of C = 11.79 / 12.0107 = 0.9816</u>

% of Cl = 69.57

Molar mass of Cl = 35.453 g/mol

<u>% moles of Cl = 69.57 / 35.453 = 1.9623</u>

Given that the gaseous chlorofluorocarbon only contains chlorine, flourine and carbon. So,

% of F = 100% - % of C - % of C = 100 - 11.79 - 69.57 = 18.64

Molar mass of F = 18.998 g/mol

<u>% moles of F = 18.64 / 18.998 = 0.9812</u>

Taking the simplest ratio for C, Cl and F as:

0.9816 : 1.9623 : 0.9812

= 1 : 2 : 1

The empirical formula is = CCl_2F

Also, Given that:

Pressure = 21.3 mm Hg

Also, P (mm Hg) = P (atm) / 760

Pressure = 21.3 / 760 = 0.02803 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K  

Volume = 458 mL  = 0.458 L (1 mL = 0.001 L)

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.02803 atm × 0.458 L = n × 0.0821 L.atm/K.mol × 298.15 K  

⇒n = 0.00052445 moles

Given that :  

Amount  = 0.107 g  

Molar mass = ?

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.00052445= \frac{0.107\ g}{Molar\ mass}

Molar\ mass= 204.0233\ g/mol

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 1×12.0107 + 2×35.453 + 1×18.998 = 101.9147 g/mol

Molar mass = 204.0233 g/mol

So,  

Molecular mass = n × Empirical mass

204.0233 = n × 101.9147

⇒ n = 2

<u>The molecular formula = C_2Cl_{4}F_2</u>

6 0
3 years ago
The maximum amounts of lead and copper allowed in drinking water are 0.015 mg/kg for lead and 1.3 mg/kg for copper. Tell the max
GuDViN [60]

Answer:

  • <em>The maximum amount of copper allowed in 100 g of water is </em><u><em>0.00013 g</em></u>

Explanation:

To find the maximum amount of copper (in grams) allowed in 100 g of water use the maximum amount ratio (1.3 mg / kg)  and set a proportion with the unknown amount of copper (x) and the amount of water (100 g):

First, convert 100 g of water to kg: 100 g × 1 kg / 1000 g = 0.1 kg.

Now, set the proportion:

  • 1.3 mg Cu / 1 Kg H₂O = x / 0.1 kg H₂O

Solve for x:

  • x = 0.1 kg H₂O × 1.3 mg Cu / 1 kg H₂O = 0.13 mg Cu

Convert mg to grams:

  • 0.13 mg × 1 g / 1,000 mg = 0.00013 g

Answer: 0.00013 g of copper.

6 0
2 years ago
Helllppppppppp Don’t Guess
ruslelena [56]

Answer:

C. Lithium

Explanation:

I Goo gled it and I think that's right.

3 0
2 years ago
How many calories are required to raise the temperature of 75g of water from 20 ˚C to 50˚ C?
mojhsa [17]
The answer to the question is letter b
7 0
3 years ago
what is the specific heat of a substance if 1560 cal are required to raise the temperature of a 312-g sample by 15°C
jasenka [17]

Answer:

0.33 cal⋅g-1°C-1  

Explanation:

The amount of heat required is determined from the formula:

q= mcΔT

To see more:

https://api-project-1022638073839.appspot.com/questions/what-is-the-specific-heat-of-a-substance-if-1560-cal-are-required-to-raise-the-t#235434

7 0
2 years ago
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