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Oksanka [162]
3 years ago
15

Why does food wrapped in aluminum foil keep warm for a long time? A) Aluminum foil is an insulator and prevents heat loss by con

duction. B) Aluminum foil forms an air-tight seal and prevents heat loss by convection. C) The silver surface of the foil reflects heat in and prevents heat loss by radiation. D) The silver surface of the foil absorbs heat from the surroundings and keeps the food warm.
Chemistry
2 answers:
vlabodo [156]3 years ago
8 0

the correct answer is c.


denis-greek [22]3 years ago
3 0

The answer is A or D  

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Which describes the volume of 1 mol of gas at standard temperature and pressure?
Setler79 [48]

Volume of 1  mol of gas at standard temperature and pressure is 22.4 L.

That is using ideal gas equation:

PV = nRT

P=pressure

V=volume

n=number of moles

R=gas constant

T=temperature

at STP,

P=1 atm

T=273K

n=1(given)

Putting all the values in the equation will give,

V= 22.4 L

So, the answer is :

The volume of 1 mol of gas at standard temperature and pressure is 22.4 L.


3 0
3 years ago
What kind of scientist would study how mistletoe harms trees?
posledela

Answer:

Biologist

Explanation:

5 0
3 years ago
for the reaction HCI+NaOH➡️NaCi+H2O, how many moles of hydrochloric acid are required to produce 150 g. of water?
Ne4ueva [31]

Answer: 8.33 mol of HCl (Hydrochloric Acid)

Explanation:

150 g H2O  x __1 mol__ x __1 mol HCl__  =  8.33 mol of HCl

                         18.016 g        1 mol H2O

6 0
2 years ago
It takes 11.2 kj of energy to raise the temperature of 145 g of benzene from 22.0°c to 67.0°c. what is the specific heat of benz
Cloud [144]
We can use the heat equation,
Q = mcΔT 

where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).
Q = 11.2 kJ = 11200 J
m = <span>145 g
</span>c = ?
ΔT = (67 - 22) °C = 45 °C
By applying the formula,
11200 J = 145 g x c x 45 °C
           c = 1.72 J g⁻¹ °C⁻¹

Hence, specific heat of benzene is 1.72 J g⁻¹ °C⁻¹.
7 0
3 years ago
There are two common oxides of copper; one is 80% copper and the other is 89% copper by weight. Calculate the formula and name e
irakobra [83]

80% copper (Cu)

Cu: 80 :  63.546 = 1.259

O: 20 : 16 = 1.25

Cu:O = 1 : 1

the formula: CuO

89% copper (Cu)

Cu: 89 :  63.546 = 1.4

O: 11 : 16 = 0.6875

Cu:O = 2:1

the formula: Cu₂O

8 0
2 years ago
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