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2 years ago

1/2 * m * v ^ 2 = mgh

Chemistry

1 answer:

viktelen [127]2 years ago

4 0

C. the square root of 2gh

Explanation:

The square root of 2gh is equivalent to Torricelli's law in the equation.

given expression:

$\frac{1}{2}$ m v² = mgh

to find v, we make it the subject of the expression:

*multiply both sides of the equation by 2:

2 x( $\frac{1}{2}$ m v² )= 2 (mgh)

mv² = 2mgh

*cancel the mass, m appearing on both sides:

v² = 2gh

* find the square root of both side

v = √2gh

Learn more:

Expression brainly.com/question/11207748

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I really need the answers please!

krek1111 [17]

1) D = 13.6 g / mL

2)ethyl alcohol weighs 158g

3)ρ _copper = 8.9 g $cm^{3}$

Explanation:

D = m / V

=306.0 g / 22.5 mL

D= 13.6 g / mL

density = mass / volume

mass = density × volume

=0.789g /ml × 200.0 ml

M=158g

Ethyl alcohol weighs 158g

ρ (density) = Mass / Volume

ρ _copper = 1896 g / 8.4cm × 5.5cm × 4.6cm

= 1896g / 212.5 $cm^{3}$

ρ _copper=8.9 g $cm^{3}$

4 0

2 years ago

If the starting volume of a hot air balloon is 55,500 m3and the initial temperature is 21 °C, what is the temperature inside the

Dmitry_Shevchenko [17]

Answer:

T₂ = 392 K

Explanation:

Given that,

Initial volume of the hot air balloon, V₁ = 55500 m³

Initial temperature, T₁ = 21°C = 294 K

Final volume, V₂ = 74000 m³

We need to find the final temperature inside the balloon. The relation between the temperature and volume is given by charles law i.e.

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

Where

T₂ is the final temperature

So,

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\T_2=\dfrac{T_1V_2}{V_1}\\\\T_2=\dfrac{294\times 74000 }{55500 }\\\\T_2=392\ K$

So, the new temperature is 392 K.

8 0

2 years ago

How many moles of KBr are present in 500 ml of a 0.8 M KBr solution?

faltersainse [42]

Answer:

2) 0.4 mol

Explanation:

Step 1: Given data

Volume of the solution (V): 500 mL

Molar concentration of the solution (M): 0.8 M = 0.8 mol/L

Step 2: Convert "V" to L

We will use the conversion factor 1 L = 1000 mL.

500 mL × 1 L/1000 mL = 0.500 L

Step 3: Calculate the moles of KBr (solute)

The molarity is the quotient between the moles of solute (n) and the liters of solution.

M = n/V

n = M × V

n = 0.8 mol/L × 0.500 L = 0.4 mol

4 0

2 years ago

Calculate ΔH∘f for NO(g) at 435 K, assuming that the heat capacities of reactants and products are constant over the temperature

weeeeeb [17]

Answer:

91383 J

Explanation:

The equation of the reaction can be represented as:

$\frac{1}{2} N_{2(g)}+\frac{1}{2} O_{2(g)}$ ------> $NO_{(g)}$

Given that:

The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.

$\delta H^0__{R,T_2}$ = $\delta H^0__{R,T_1} } + \int\limits^{T_2}_{T_1} {\delta C_p(T')} \, dT'$

where:

$\delta H^0__{R}$ = enthalpy of reaction

${\delta C_p(T')}$ = the difference in the heat capacities of the products and the reactants.

∴

$\delta H^0__{R,435K}$ = $\delta H^0__{R,298.15K}$ $+$ $\int\limits^{435}_{298.15} {\delta C_p(T')} \, dT'$

= $1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'$

= 91300 J + (0.605 J.K⁻¹)(435-298.15)K

= 91382.79 J

$\delta H^0__{R,435K}$ ≅ 91383 J

6 0

3 years ago

Name the fundamental unit involved in the derived unit joule?

tangare [24]

Answer:

Energy

Explanation:

7 0

2 years ago