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3 years ago

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Which of the following would happen if Earth's moon were half the size that it is now? O The oceans' tidal variations would be s

maller because the Moon would exert less gravitational pull on Earth's oceans. O The oceans' tidal variations would be greater because the Moon would exert less gravitational pull on Earth's oceans. O The oceans' tidal variations would be greater because the Moon would exert more gravitational pull on Earth's oceans. The oceans' tidal variations would be smaller because the Moon would exert more gravitational pull on Earth's oceans. & Previous

1 answer:

Brilliant_brown [7]3 years ago

3 0

Answer: A. The oceans‘ tidal would be smaller because the moon would exert less gravitational pull on earths oceans.

Explanation:

i got it right :)

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The standard Gibbs free energy of formation of ________ is zero. (

zvonat [6]

The standard Gibbs free energy of formation ΔGf° of Rb(s), H2(g) and Pb(s) are all zero. Similar to enthalpies of formation, the values of the standard Gibbs energies of formation are zero for the elements in their most stable forms at room conditions 298 Kelvin and one atmosphere pressure.

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3 years ago

NH3 + HCl---> NH4Cl What is the Chemical reaction

KATRIN_1 [288]

Answer:

A combination reaction

Explanation:

The chemical reaction between ammonia and hydrochloric acid as shown below:

NH₃ + HCl → NH₄Cl

is a combination reaction.

In a combination reaction, two compounds combines together to give one compound.

A combination reaction is also known as a synthesis reaction.

A single product forms from tow or more reactants.

The driving force for such reaction is the large and negative heat of formation of the product.

5 0

3 years ago

what grade is learning What amount of thermal energy (heat) in joules required to raise the temperature of 25 grams of water fro

blsea [12.9K]

Answer:

10425 J are required

Explanation:

assuming that the water is entirely at liquid state at the beginning , the amount required is

Q= m*c*(T final - T initial)

where

m= mass of water = 25 g

T final = final temperature of water = 100°C

T initial= initial temperature of water = 0°C

c= specific heat capacities of water = 1 cal /g°C= 4.186 J/g°C ( we assume that is constant during the entire temperature range)

Q= heat required

therefore

Q= m*c*(T final - T initial)= 25 g * 4.186 J/g°C * (100°C- 0°C) = 10425 J

thus 10425 J are required

7 0

4 years ago

Which of the following statements is true of enzymes A. They're unusable after a reaction B. They don't provide an active site f

Mnenie [13.5K]

D. They act on a specific type of substrate in a reaction

Hope this helps!

3 0

3 years ago

Read 2 more answers

A compound contains only carbon, hydrogen, and oxygen. combustion of 11.75 mg of the compound yields 17.61 mg co2 and 4.81 mg h2

Ymorist [56]

Number of moles is defined as the ratio of given mass in g to the molar mass.

First, convert the given mass of carbon dioxide in mg to g:

1 mg = 0.001 g

17.61 mg = 0.01761 g

Number of moles of carbon dioxide = $\frac{0.01761 g}{44.01 g/mol}$

= $0.0004001 mol$

Mass of carbon = number of moles of carbon dioxide \times molar mass of carbon

= $0.0004001 mol\times 12.011 g/mol$

= $0.004806 g$

Number of moles of water= $\frac{0.00481 g}{18 g/mol}$

= $2.672\times 10^{-4}$

Since, water contains two hydrogen atoms. Thus,

Moles of hydrogen = $2\times 2.672\times 10^{-4}$

= $5.34\times 10^{-4}$

Mass of hydrogen = $5.34\times 10^{-4}\times \times 1.008 g/mol$

= $5.34\times 10^{-4} g$

Mass of oxygen = $0.001175-(5.38\times 10^{-4}g+0.004806 g)$

= $0.006405 g$

Number of moles of oxygen = $\frac{0.006405 g}{15.999 g/mol}$

= $0.000400$

Now,

$C_{0.0004001} H_{0.000534} O_{0.000400}$

Divide the smallest number to get the whole number,

$C_{\frac{0.0004001}{0.000400}} H_{\frac{0.000534}{0.000400}} O_{\frac{0.000400}{0.000400}}$

we get,

$C_{1} H_{1.33} O_{1}$

Now, multiply all the subscript by 3 to get the whole number,

$C_{3} H_{4} O_{3}$ (empirical fomula)

Molar mass of the compound = $3\times 12.011 g/mol+4\times 1.008 g/mol+3\times 15.999 g/mol$

= $88.062 g/mol$

Divide given molar mass of the compound with the molar mass of the compound.

= $\frac{176.1 g/mol}{88.062 g/mol}$

= $1.999\simeq 2$

Thus, multiply the subscripts of empirical formula by 2 to get the molecular formula, we get:

$C_{6}H_{8}O_{6}$

Hence, empirical formula is $C_{3}H_{4}O_{3}$ and molecular formula is $C_{6}H_{8}O_{6}$

8 0

3 years ago