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SCORPION-xisa [38]
3 years ago
8

The pot shown in the picture is made of two different metals. Why is the handle made of a different metal than the rest of the p

ot?
Chemistry
2 answers:
sertanlavr [38]3 years ago
8 0
The rest of the bot needs to be able to heat up to cook ur food but the handles should be cooler so u can touch them
scoray [572]3 years ago
6 0

D) The metal in the handle can absorb more energy without changing temperature as easily as the metal in the pot. I know because I just answered this problem.

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What is the percent by volume of a solution formed by mixing 15 mL of hydrochloric acid into water, forming 75 mL of solution?
timama [110]

Answer:

= 17% (v/v)  

Explanation:

% by Volume = (Volume HCl / Volume of Solution)100%

= 15ml/(75ml + 15ml) x 100% = 16.66666667% (calculator answer)

= 17% (v/v)  2 sig figs. based on volume values given.

4 0
3 years ago
Who turned the ragged Continental Army into a more efficient fighting force?
Goshia [24]
Von Steuben turned the ragged Continental Army into a more effective fighting force by drilling and teaching the army discipline
8 0
3 years ago
A runner wants to run 10.4 km k m . She knows that her running pace is 8.0 mi/h m i / h .
mars1129 [50]

Answer:

Approximately 49 minutes.

Explanation:

1 mile = 1.6 km

So 8.0 x 1.6 km

Minutes required: \frac{10.4 times 60 }{8.0 times 1.6}= 48.75 minutes

I hope this helped :)

4 0
3 years ago
What is the ratio of acetate ion to acetic acid (Ka = 1.76×10–5) in a solution containing these compounds at pH 4.38 ? Write you
jenyasd209 [6]

Answer:

0.43.

Explanation:

In order to solve this question we will need the an equation called Henderson-Hasselbalch Equation. The Henderson-Hasselbalch Equation can be represented by the reaction below;

pH= pKa + log ( [ A^- ] / [ HA] ).

Where HA is the acetic acid and A^- is the Acetate ion

We are given the pH value to be = 4.38 and the ka to be = 1.76×10^–5. So, we will use the value for the ka to find the pKa through the formula below.

pKa = - log ka.

Therefore, pKa = - log( 1.76×10^–5).

pKa= 4.75 + log

So,

4.38 = 4.75 + log ([ A^-] / [HA]).

4.38 - 4.75 = log ( [ A^- ] / [ HA] ).

( [ A^- ] / [ HA] ) = 10^- 0.37.

( [ A^- ] / [ HA] ) = 0.42657951880159265.

( [ A^- ] / [ HA] )= 0.43.

6 0
3 years ago
How many grams of carbon dioxide are created from the complete combustion of 21.3 L of butane at STP?
Bond [772]

Answer:

165.5 g of CO2

Explanation:

We must first put down the balanced reaction equation:

C4H10(g) + 13/2 O2(g) ------> 4CO2(g) + 5H2O(g)

From the reaction equation, one mole of butane occupies 22.4 L hence we can establish the stoichiometry of the reaction thus:

22.4 L of butane created 174 g of CO2

Therefore 21.3 L of butane will create 21.3 × 174/22.4 = 165.5 g of CO2

3 0
4 years ago
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