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3 years ago

Foods that allow microorganism to grow are called parasites true or false

1 answer:

7 0

False. Foods that allow microorganism to grow are not called parasites. Parasites are organisms that feeds on the nutrients of its host.
For example leeches. They suck on our blood.

Foods that allow microorganism to grow are called potentially hazardous foods. They are high in protein and moisture and are slightly acidic.
For example hamburgers. They are high in protein and moisture.

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Aluminum chloride can be formed from its elements:

saul85 [17]

Answer: The $\Deltas H^o_{formation}$ for the reaction is -1406.8 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical reaction for the formation reaction of $AlCl_3$ is:

$2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3 (s)$ $\Delta H^o_{formation}=?$

The intermediate balanced chemical reaction are:

(1) $HCl(g)\rightarrow HCl(aq.)$ $\Delta H_1=-74.8kJ$ ( × 6)

(2) $H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$ $\Delta H_2=-185kJ$ ( × 3)

(3) $AlCl_3(aq.)\rightarrow AlCl_3(s)$ $\Delta H_3=+323kJ$ ( × 2)

(4) $2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)$ $\Delta H_4=-1049kJ$

The expression for enthalpy of formation of $AlCl_3$ is,

$\Delta H^o_{formation}=[6\times \Delta H_1]+[3\times \Delta H_2]+[2\times \Delta H_3]+[1\times \Delta H_4]$

Putting values in above equation, we get:

$\Delta H^o_{formation}=[(-74.8\times 6)+(-185\times 3)+(323\times 2)+(-1049\times 1)]=-1406.8kJ$

Hence, the $\Deltas H^o_{formation}$ for the reaction is -1406.8 kJ.

6 0

3 years ago

How many total moles of ions are released when the following sample dissolves completely in water? Enter your answer in scientif

Katena32 [7]

Answer: The number of moles of strontium bicarbonate is $7.5\times 10^{-9}mol$

Explanation:

Formula units are defined as lowest whole number ratio of ions in an ionic compound. It is calculate by multiplying the number of moles by Avogadro's number which is $6.022\times 10^{23}$

We are given:

Number of formula units of $Sr(HCO_3)_2=4.55\times 10^{15}$

As, $6.022\times 10^{23}$ number of formula units are contained in 1 mole of a substance.

So, $4.55\times 10^{15}$ number of formula units will be contained in = $\frac{1}{6.022\times 10^{23}}\times 4.55\times 10^{15}=7.5\times 10^{-9}mol$ of strontium bicarbonate.

Hence, the number of moles of strontium bicarbonate is $7.5\times 10^{-9}mol$

5 0

3 years ago

Ce cantitate de O se gaseste in 5,8 g hidroxid de magneziu

coldgirl [10]

Answer:

separare pâlnii de picurare balon cotat. 1. 2. 3. 4 5 6 7. 8. 9. 1 ... Mod de lucru: 25 g Na2SO4∙7H2O se dizolvă în cantitatea minim ... Exemple: NaOH – hidroxid de sodiu.

Explanation:

make me as brain liest

4 0

4 years ago

The stomach and intestines are organs of the body system which _______.

yulyashka [42]

Answer:

A. Digests food

Explanation:

Look about diggestive process in Google

Best regards

3 0

3 years ago

Read 2 more answers

When 16.0 g of an unknown compound (a nonelectrolyte) are dis solved in exactly 800. g of water, the solution has a freezing poi

dexar [7]

Answer:

A. 266g/mol

Explanation:

A colligative property of matter is freezing point depression. The formula is:

ΔT = i×Kf×m (1)

Where:

ΔT is change in temperature (0°C - -0,14°C = 0,14°C)i is Van't Hoff factor (1 for a nonelectrolyte dissolved in water), kf is freezing point molar constant of solvent (1,86°Cm⁻¹) and m is molality of the solution (moles of solute per kg of solution). The mass of the solution is 816,0g

Replacing in (1):

0,14°C = 1×1,86°Cm⁻¹× mol Solute / 0,816kg

0,0614 = mol of solute.

As molar mass is defined as grams per mole of substance and the compound weights 16,0g:

16,0g / 0,0614 mol = 261 g/mol ≈ A. 266g/mol

I hope it helps!

3 0

3 years ago