Can I use the sap of jackfruit to be used into a glue?

A flashlight has a resistance of 2.4 . What voltage is applied by the batteries if the current in the circuit is

hammer [34]

Answer : The voltage applied by the batteries is, 6.0 V

Solution : Given,

Resistance of flashlight = 2.4 ohm

Current in the circuit = 2.5 Ampere

Formula used :

$V=IR$

where,

V = applied voltage

I = current in the circuit

R = resistance of light

Now put all the given values in the above formula, we get

$V=(2.5A)\times (2.4ohm)=6volt=6V$

Therefore, the voltage applied by the batteries is, 6.0 V

6 0

3 years ago

Read 2 more answers

Ices Creamery combines sugar, cream, and the best natural flavorings, that churns and freezes the resulting mixture at a very lo

Rasek [7]

Answer:

Process manufacturing.

Explanation:

Process manufacturing, a class of manufacturing which is associated with the formulas and the manufacturing of recipes. This type of manufacturing is different from discrete manufacturing as it deals with discrete units and assembly of the components.

<u> This type of manufacturing is common in food, chemical, beverage, nutraceutical, pharmaceutical, cannabis, and mostly biotechnology industries. In this manufacturing, the relevant factors are the ingredients and the bulk materials rather than the individual units. </u>

<u>Thus, formation of ice is an example of process manufacturing in which different ingredients are added and using recipe of freezing.</u>

7 0

3 years ago

A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (

djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

$Kb = \frac{[ HSO3-][0H-]}{SO3-2}$

$Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7$

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation $Kb=\frac{[HSO3-][OH-]}{[SO3-2]}$

We noe have:

$1.485*10^-^7=\frac{x*x}{(0.056-x)}$

$x = [OH-] = 9.11*10^-^5$

$pOH = -log(OH) = -log(9.11*10^-^5)$

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0

3 years ago

CH4+2 O2-> CO2 + 2 H2O

Elenna [48]

The right answer is the last one hope this helps your very welcome

8 0

3 years ago

introduced in the 13th-16th? century from sabah in the reign of the sultanate of sulu, and became extinct on maguindanao or were

arsen [322]

The Elephant Maximus is the species that was introduced and later became extinct on Maguindanao.

Extinction occurs as all members of a species die, and therefore the species disappears. This includes complete extinction or extinction of the species in a specific zone.

In the case of the Elephant Maximus, which is known as the Asian Elephant, this species became extinct in a region known as Maguindanao in the Philippines.

This partial extinction occurred as the elephants from this zone were taken to Sabah in Malaysia from the 13th century to the 16th century. This occurred because the Sultans of this region considered this elephant to be a valuable and even sacred animal.

Learn more in: brainly.com/question/14698336

4 0

3 years ago