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4 years ago

How much time will it take a car travelling at 88 km/hr (55mi/hr)to travel 500km?

1 answer:

7 0

The time it will take for a car travelling 88 km/hr (55mi/hr) to travel 500km would be 3.5 hours.

How? Add 88 + 58.

That'll equal 143, then divide 500 ÷ 143 = 3.5

So, 3.5 is the time that the car will take travelling 88 km/hr trying to reach the destination of 500km.

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PLEASEEEEEEEEEEEE HELPPPPPPPP I BEGGGGG FOR HELPPPPP

Elza [17]

Answer: There are $21.08\times 10^{23}$ molecules in 63.00 g of $H_2O$

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{63.00g}{18g/mol}=3.5moles$

1 mole of $H_2O$ contains = $6.023\times 10^{23}$ molecules

Thus 3.5 moles of $H_2O$ contains = $\frac{6.023\times 10^{23}}{1}\times 3.5=21.08\times 10^{23}$ molecules.

There are $21.08\times 10^{23}$ molecules in 63.00 g of $H_2O$

3 0

3 years ago

Write the formula for Zn+2 (SO4)-2

Pie

Answer:

Zn2 + SO4 = Zn2(SO4)2

Explanation:

4 0

3 years ago

Oxalic Acid, a compound found in plants and vegetables such as rhubarb, has a mass percent composition of 26.7% C, 2.24% H, and

blondinia [14]

Answer:

HCO₂

Explanation:

From the information given:

The mass of the elements are:

Carbon C = 26.7 g; Hydrogen H = 2.24 g Oxygen O = 71.1 g

To determine the empirical formula;

First thing is to find the numbers of moles of each atom.

For Carbon:

$=26.7 \ g\times \dfrac{1 \ mol }{12.01 \ g} \\ \\ =2.22 \ mol \ of \ Carbon$

For Hydrogen:

$=2.24 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =2.22 \ mol \ of \ Hydrogen$

For Oxygen:

$=71.1 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =4.44 \ mol \ of \ oxygen$

Now; we use the smallest no of moles to divide the respective moles from above.

For carbon:

$\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ carbon$

For Hydrogen:

$\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ hydrogen$

For Oxygen:

$\dfrac{4.44 \ mol \ of \ Oxygen}{2.22} =2 \ mol \ of \ oxygen$

Thus, the empirical formula is HCO₂

4 0

3 years ago

Copper has a density of 8.9 g cm3. if you have 90 g of copper how much volume is your occupying?

babunello [35]

8.9 g + 90 g = 98.9 Copper has a density

4 0

3 years ago

The temperature of a 100.0 g sample of water is raised from 30degrees celsius to 100.0 degrees celsius. How much energy is requi

worty [1.4K]

Answer:

29260J

Explanation:

Given parameters:

Mass of water sample = 100g

Initial temperature = 30°C

Final temperature = 100°C

Unknown:

Energy required for the temperature change = ?

Solution:

The amount of heat required for this temperature change can be derived from the expression below;

H = m c (ΔT)

H is the amount of heat energy

m is the mass

c is the specific heat capacity of water = 4.18J/g°C

ΔT is the change in temperature

Now insert the parameters and solve;

H = 100 x 4.18 x (100 - 30)

H = 100 x 4.18 x 70 = 29260J

6 0

3 years ago