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3 years ago

This is a science question. I have two of them.

1 answer:

7 0

Second Question: An endospore is a dormant, tough, and non-reproductive structure produced by certain bacteria from the Firmicute phylum. Endospore formation is usually triggered by lack of nutrients, and usually occurs in Gram-positive bacteria. In endospore formation, the bacterium divides within its cell wall.

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Find the volume of 3.011x1023 molecules of O2 at STP.

lord [1]

Answer:

6.022 x 1023 molecules = 1 Mole O2 = 22.4L

3.022 x 1023 molecules = 1/2 Mole O2 = 11.2L

The answer is 11.2 Litres

Explanation:

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2 years ago

What causes clothes from a dryer to stick together?

leva [86]

Different fabrics rub together, and electrons may rub off

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3 years ago

Write the structure of 2 butene class 10

prohojiy [21]

Answer:

The answer is C4H8.

Explanation:

I hope it helps you

Pass your test fast okay

BYE!

3 0

4 years ago

Compounds in alcoholic beverages that enhance flavor and appearance but may contribute to hangover symptoms are called:_____.

olga_2 [115]

Compounds in alcoholic beverages that enhance flavor and appearance but may contribute to hangover symptoms are called congeners.

<h3>Alcoholic beverages:</h3>

Congeners are compounds that add to the flavor, smell, and appearance of most alcoholic beverages. These substances may make hangover symptoms worse. Because they contain fewer congeners than whiskey, brandy, and red wine, clear alcoholic beverages like gin and vodka have less of a hangover-inducing effect.

The impact of ethanol, or the alcohol in your drinks, is the primary contributor to a hangover. It is a poisonous substance that acts as a diuretic in the body, which causes you to urinate more frequently and increases the likelihood that you will become dehydrated. The incidence and intensity of hangovers are both increased by congeners, substances created during the digestion and maturation of alcohol.

Learn more about congeners here:

brainly.com/question/1837839

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3 0

2 years ago

A student prepared an equilibrium solution by mixing the following solutions:A.2.00 mL of 0.00250 M Fe(NO3)3B.5.00 mL of 0.00250

Crank

Explanation:

$Molarity=\frac{\text{Moles of solute}}{\text{Volume of solution Liter}}$

A. 2.00 mL of 0.00250 M $Fe(NO_3)_3$

Moles of ferric nitrate = n

Volume of ferric nitrate = 2.00 ml = 0.002 L ( 1 mL=0.001 L)

Molarity of ferric nitrate = 0.00250 M

$n=0.00250 M\times 0.002 L=0.000005 mol$

B. 5.00 mL of 0.00250 M $KSCN$

Moles of KSCN = n'

Volume of KSCN = 5.00 ml = 0.005 L ( 1 mL=0.001 L)

Molarity of KSCN = 0.00250 M

$n'=0.00250 M\times 0.005 L=0.0000125 mol$

C. 3.00 mL of 0.050 M $HNO_3$

Moles of nitric acid = n''

Volume of nitric acid = 3.00 ml = 0.003 L ( 1 mL=0.001 L)

Molarity of nitric acid = 0.050 M

$n=0.050 M\times 0.003 L=0.00015 mol$

After mixing A, B and C together and their respective initial concentration before reaction.

After mixing A, B and C together the volume of the solution becomes = V

V = 0.002 L=0.005 L+0.003 L= 0.010 L

Concentration of ferric nitrate :

$[Fe(NO_3)_3]=\frac{0.000005 mol}{0.010 L}=0.0005 M$

Concentration of ferric ions :

$[Fe^{3+}]=1\times [Fe(NO_3)_3]=0.0005 M$

Concentration of nitrate ions from ferric nitrate:

$[NO_3^{-}]=3\times [Fe(NO_3)_3]=0.0015 M$

Concentration of KSCN :

$[KSCN]=\frac{0.0000125 mol}{0.010 L}=0.00125 M$

Concentration of $SCN^-$ ions:

$[SCN^-]=1\times [KSCN]=0.00125 M$

Concentration of potassium ions:

$[K^+]=1\times [KSCN]=0.00125 M$

Concentration of nitric acid :

$[HNO_3]=\frac{0.00015 mol}{0.010 L}=0.015 M$

Concentration of hydrogen ion :

$[H^+]=1\times [HNO_3]=0.015 M$

Concentration of nitrate ions from nitric acid :

$[NO_3^{-}]=1\times [HNO_3]=0.0015 M$

Concentration of nitrate ion in mixture = 0.0015 M + 0.0015 M = 0.0030 M

$Fe^{3+}+SCN^-\rightleftharpoons Fe(NCS)^{2+}$

given concentration of $Fe(NCS)^{2+}$ at equilbrium = $3.6\times 10^{-5} M = 0.000036 M$

initially :

0.0005 M 0.00125 M 0

At equilibrium

(0.0005-0.000036) M (0.00125-0.000036) M 0.000036 M

0.000464 M 0.001214 M 0.000036 M

The expression of an equilibrium constant will be given as;

$K_c=\frac{[Fe(NCS)^{2+}]}{[Fe^{3+}][SCN^{-}]}$

$=\frac{0.000036 M}{0.000464 M\times 0.001214 M}=63.91$

The value for the equilibrium constant is 63.91.

6 0

3 years ago