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3 years ago

Is utensils a substance homogeneous mixture or heterogeneous mixture

1 answer:

7 0

A pure substance or a homogeneous mixture consists of a single phase. A heterogeneous mixture consists of two or more phases. When oil and water are combined, they do not mix evenly, but instead form two separate layers.

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5.4×10^3 × 1.2×10^7

Nadya [2.5K]

5.4*10^3 * 1.2*10^7.

5.4 * 1000 * 1.2 * 10000000

Multiply them = 64800000000

4 0

3 years ago

write a complete ionic equation and a net ionic equation for the reaction of aqueous solutions of calcium nitrate and iron (III)

maw [93]

3Ca(NO3)2(aq) +Fe2(SO4)3(aq)-----> 3CaSO4(s)+ 2Fe(NO3)3(aq)
3Ca²⁺ + 6 NO3 ⁻ +2 Fe ³⁺ +3SO4²⁻ -----> 3CaSO4(s)+2 Fe ³⁺+ 6 NO3 ⁻
3Ca²⁺+3SO4²⁻---->3CaSO4(s)
Ca²⁺+SO4²⁻---->CaSO4(s)

7 0

3 years ago

What temperature change in Celsius degrees is produced when 800. J is absorbed by 100. G of water?

g100num [7]

Answer: The final temperature is equal to

45 Celsius

Explanation:

We have the amount of energy gain

⋅

where

4.184

is the specific heat of water,

is the mass of water

⇒

840

4.184

⋅

(

−

)

840

4.184

i.e.

∘

6 0

4 years ago

Which of the following equations demonstrates the law of conservation of mass?

VLD [36.1K]

Answer:

Mg + O2 ---> MgO

Explanation: V2O5 + CaS → CaO + V2S5. 5V2O5 + 5CaS → 10CaO + 5V2S5. 3V2O5 + 3CaS → 3CaO + 3V2S5.

7 0

3 years ago

7. A gas-filled weather balloon with a volume of 65.0 L is released at sea-level conditions of

Galina-37 [17]

The balloon will reach its maximum volume and it will burst.

Given:

A weather balloon at sea level, with gas at 65.0 L volume, 745 Torr pressure, and 25C temperature.
When the balloon was taken to an altitude at which temperature was 25C and pressure was 0.066atm its volume expanded.
The maximum volume of the weather balloon is 835 L.

To find:

Whether the weather balloon will reach its maximum volume or not.

Solution:

The pressure of the gas in the weather balloon at sea level = $P_1=745 torr$

$1 atm = 760 torr\\P_1=745 torr=\frac{745}{760} torr = 0.980 atm$

The volume of the weather balloon at sea level = $V_1=65.0L$

The temperature of the gas in the weather balloon at sea level:

$T_1=25^oC=25+273.15 K=298.15 K$

The balloon rises to an altitude.

The pressure of the gas in the weather balloon at the given altitude:

$P_2=0.066atm$

The volume of the weather balloon at the given altitude = $V_2=?$

The temperature of the gas in the weather balloon at the given altitude:

$T_1=25^oC=25+273.15 K=298.15 K$

Using the Combined gas law:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\frac{0.980 atm\times 65.0L}{298.15 K}=\frac{0.066atm\times V_2}{298.15K}\\V_2=\frac{0.980 atm\times 65.0L\times 298.15K}{0.066atm\times 298.15 K}\\=965L$

The maximum volume of the weather balloon= V = 835 L

$V < V_2$

The volume of the weather balloon at a given altitude is greater than its maximum volume which means the balloon will reach its maximum volume and it will burst.

Learn more about the combined gas law:

brainly.com/question/13154969?referrer=searchResults

brainly.com/question/936103?referrer=searchResults

4 0

3 years ago