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vredina [299]
3 years ago
8

2. His_________would take place tomorrow. (bury)

Chemistry
1 answer:
rusak2 [61]3 years ago
6 0

Answer:

1. burial

2. Choice

3. pleasure

Explanation:

Mark brainliest! Please!

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Sports drinks have two very important ingredients, what are they?
Nookie1986 [14]
Whenever a sports person engages in a physical activity, there is a lot of sweat. The sweat causes loss of water and electrolytes in the body. So, both need to be restored back to the required amount in the body.

ELECTROLYTES AND WATER.
4 0
3 years ago
A sample of gas in a balloon has an initial temperature of 21 ∘C and a volume of 1310 L . If the temperature changes to 70 ∘C ,
bixtya [17]

Answer:

1528.3L

Explanation:

To solve this problem we should know this formula:

V₁ / T₁ = V₂ / T₂

We must convert the values of T° to Absolute T° (T° in K)

21°C + 273 = 294K

70°C + 273 = 343K

Now we can replace the data

1310L / 294K = V₂ / 343K

V₂ = (1310L / 294K) . 343K → 1528.3L

If the pressure keeps on constant, volume is modified directly proportional to absolute temperature. As T° has increased, the volume increased too

7 0
4 years ago
Elements located in the vertical column of the periodic table are of the same and have____chemical properties.
umka2103 [35]

Answer:

Similar

Explanation:

Elements that are located in the same vertical colomn or group of the periodic table have similar chemical properties. This is because they have the same amount of valence electrons in the outermost shell.

4 0
3 years ago
A research balloon at ground level contains 12 L of helium (He) at a pressure of 725mmHg and a temperature of 30.00∘C. When the
Helga [31]

<u>Answer:</u> The temperature when the volume and pressure has changed is -27.26°C

<u>Explanation:</u>

To calculate the temperature when pressure and volume has changed, we use the equation given by combined gas law. The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1,V_1\text{ and }T_1 are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2 are the final pressure, volume and temperature of the gas

We are given:

P_1=725mmHg\\V_1=12L\\T_1=30.00^oC=[30+273]K=303K\\P_2=252mmHg\\V_2=28L\\T_2=?K

Putting values in above equation, we get:

\frac{725mmHg\times 12L}{303K}=\frac{252mmHg\times 28L}{T_2}\\\\T_2=\frac{252\times 28\times 303}{725\times 12}=245.74K

Converting this into degree Celsius, we get:

T(K)=T(^oC)+273

245.74=T(^oC)+273\\T(^oC)=-27.26^oC

Hence, the temperature when the volume and pressure has changed is -27.26°C

5 0
3 years ago
Suppose a 20.0 g gold bar at 35.0°C absorbs 70.0 calories of heat energy. Given that the specific heat of gold is 0.0310 cal/g °
timofeeve [1]

We know, change in temperature is given by :

T_2-T_1=\dfrac{q}{mC_p(Gold)}

Putting all given values, we get :

T_2-T_1=\dfrac{70\ cal}{20\ g\times 0.0310\ cal/g^o\ C}\\\\T_2-T_1=112.90^oC\\\\T_2-35^oC=112.90^oC\\\\T_2=(112.90+35)^oC\\\\T_2=147.9^oC

Hence, this is the required solution.

6 0
4 years ago
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