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2 years ago

Cell Walls are found in Prokaryotes and Plant cells. True O False

Chemistry

2 answers:

Natasha2012 [34]2 years ago

5 0

The answer is true. Cell walls are found in both. :)

insens350 [35]2 years ago

4 0

Answer:

i think true

Explanation:

lemme know if right if not sorry :))

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Explain why the nucleus of an atom has a positive electrical charge.

xxTIMURxx [149]

Answer: Nucleus of an atom is heavy and positively charged because nucleus is only made up of protons and neutrons.

6 0

3 years ago

How many grams are needed to prepare 750. mL of a 0.35 M solution of NaOH?

nikdorinn [45]

Answer:

10.5grams

Explanation:

Molarity = number of moles (n)/ volume (V)

According to this question;

Volume = 750 mL = 750/1000 = 0.75L

Molarity = 0.35M

number of moles (n) = molarity × volume

n = 0.35 × 0.75

n = 0.2625mol

Using mole = mass/molar mass

Molar Mass of NaOH = 23 + 16 + 1

= 40g/mol

mole = mass/molar mass

0.2625 = mass/40

mass = 10.5grams

10.5 grams are needed to prepare 0.75L of a 0.35 M solution of NaOH.

3 0

3 years ago

Calculate the molarity of a KCl solution if 37.3 g of KCl are dissolved in water to give a

saw5 [17]

Answer:

Explanation:

find concentration in g/cm^3,then molar mass of kcl(39+35.5),if 1mol=74.5g what about 74.6

8 0

3 years ago

List the 5 signs of a CHEMICAL REACTION

SVETLANKA909090 [29]

Color change, formation of a precipitation, formation of a gas, odor change, temperature change

6 0

3 years ago

Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago

Cell Walls are found in Prokaryotes and Plant cells. True O False​

Cell Walls are found in Prokaryotes and Plant cells. True O False