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stiv31 [10]
3 years ago
8

Get me the answers please

Chemistry
2 answers:
melomori [17]3 years ago
7 0

Answer:

a. opinion

Explanation:

shutvik [7]3 years ago
3 0

Answer:

a is a opinon answer

Explanation:

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A frying pan is connected to a 110-volt circuit. If the resistance of the frying pan is 10 ohms, how many amperes does the fryin
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<u>11 A</u>

Explanation:

⇒ V = IR

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⇒ I = <u>11 A</u>

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3 years ago
Alyssa is trying to balance the following equation: Pb(OH)2 + 2H2SO4 =&gt; PbSO4 + 2H20 What number should be used to replace th
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Answer:

2 Pb(OH)2 + 2H2SO4 => 2 PbSO4 + 4 H20

Explanation:

Since there's no "?" shown in the equation, let's balance it and solve it entirely.

Pb(OH)2 + 2H2SO4 => PbSO4 + 2H20

1Pb + 10O + 6H + 2S ≠ 1Pb + 6O + 4H + 1S → it needs to be balanced.

To do this, let's start by looking at the elements that are only presnet once on each side:

On the products half, S is only present in PbSO4 → if we look at the reagents half, we can see it needs a "2" → then Pb is multiplied by 2 too → so Pb(OH)2 on the reagents half will also need a "2" → final count on O and H on the reagents side: 12O and 8H  → to balance it, you need 4 water molecules on the products side.

6 0
3 years ago
How many calories of heat are required to raise the temperature of 1.00kg of water from 10.2 degrees Celsius to 26.8 degrees Cel
Vika [28.1K]

Answer:

Explanation:

we know that specific heat is the amount of heat required to raise the  temperature of substance by one degree mathmeticaly

Q=mcΔT

ΔT=T2-T1

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C for water is 4.184

therefore

Q=1.00*4.184*16.6

Q=69.4 j

now we have to covert joule into calorie

1 calorie =4.2 j

x calorie=69.4 j/2

so 69.4 j =34.7 calorie thats why 34.7 calorie heat is required to raise the temperature of water from 10.2 to 26.8 degree celsius

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