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4 years ago

A golf ball is resting on the grass what must happen to cause the ball to move?

Chemistry

1 answer:

eimsori [14]4 years ago

6 0

There has to be a force in order for it to move

In other words every action needs a reaction and vice versa

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12500 J/ (106 g) (4C) reduce to one

Sloan [31]

The given expression is $\frac{12500 J}{(106g)(4^{0}C )}$

This expression denotes the specific heat of a substance which indicates a heat energy of 12500 J is involved in raising the temperature of 106 g of the substance by $4^{0}C$ . Generally, the units of specific heat are $\frac{J}{g.^{0}C }$

$\frac{12500 J}{(106g)(4^{0}C )}$ = $\frac{12500 J}{(106g)(4^{0}C } = 29.5 \frac{J}{g^{0}C }$

Therefore, $\frac{12500 J}{(106g)(4^{0}C )}$ when reduced to one unit will be 29.5 $\frac{J}{g^{0}C }$

5 0

3 years ago

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Which of the following is soluble in water?

Nina [5.8K]

I hope this answers helps it’s D

8 0

3 years ago

Sodium bicarbonate is the chemical name for baking powder. Its chemical formula is NaHCO3. What does the subscript 3 mean?

Brut [27]

Answer:

Explanation:

3 0

4 years ago

24.00 mL of a 0.25 M NaOH solution is titrated with 0.10M

Goryan [66]

Okay so, 24.00 mL of a 0.25 M NaOH solution is titrated.

4 0

3 years ago

A gas mixture with a total pressure of 750 mmHg contains each of the following gases at the indicated partial pressures: CO2 , 1

jolli1 [7]

Answer: a) 211 mm Hg

b) 0.629 grams

Explanation:

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_1+p_2+p_3+p_4$

$p_{total}$ = total pressure = 750 mmHg

$p_{CO_2}$ = 124 mm Hg

$p_{Ar}$ = 218 mm Hg

$p_{O_2}$ = 197 mm Hg

$p_{He}$ = ?

$750 mmHg=124 mm Hg+218 mm Hg+197 mm Hg+p_{He}$

$p_{He}=211mmHg$

Thus the partial pressure of the helium gas is 211 mmHg.

b) According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 211 mmHg = 0.28 atm (760mmHg=1atm)

V= Volume of the gas = 13.0 L

T= Temperature of the gas = 282 K

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas= ?

$n=\frac{PV}{RT}=\frac{0.28\times 13.0}0.0821\times 282}=0.157moles$

Mass of helium= $moles\times {\text {molar mass}}=0.157\times 4=0.629g$

Thus mass of helium gas present in a 13.0-L sample of this mixture at 282 K is 0.629 grams

8 0

3 years ago