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Leokris [45]
3 years ago
6

The amount of matter in an object is called _____. gravity mass volume weight

Chemistry
2 answers:
Leya [2.2K]3 years ago
7 0

mass - hope this helps!!

Firlakuza [10]3 years ago
4 0

The answer is Mass☺️.

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How can you prevent frequent landslides from heavy rains
GrogVix [38]

Answer: make a drainage system or make sure that the ground does not absorb the water from the rain and cause a landslide.

Explanation: To prevent frequent landlides you have to make suee there is and area for the rain to go so it does not get stuck in the mud and destabalize the mud

3 0
3 years ago
Problem Page Question A major component of gasoline is octane C8H18. When liquid octane is burned in air it reacts with oxygen O
KATRIN_1 [288]

Answer:

0.3mol C8H18

Explanation:

For this we must first look at the reaction taking place:

C8H18+O2 --> H2O + CO2

Balancing the equation we get:

2(C8H18)+25(O2) --> 18(H2O) + 16(CO2)

Form there we now need to know how many moles of Octane are needed to produce 2.4moles of H2O. The conversion is as follows:

2.4molH2O ((2mol of C8H18)/(18mol of H2O)) = 0.3mol C8H18

3 0
3 years ago
Suppose a 500.mL flask is filled with 0.40mol of N2 and 1.0mol of NO. The following reaction becomes possible:
hammer [34]

Answer:

[N₂] = 1.1M

Explanation:

Based on the chemical reaction:

N₂(g) + O₂(g) ⇄ 2 NO(g)

Equilibrium constant, K, is defined as:

K = 5.93 = [NO]² / [N₂] [O₂]

<em>Where [] are equilibrium concentrations of each specie</em>

<em />

As initial concentrations are:

N₂ = 0.40mol / 0.500L = 0.8M

NO = 1mol / 0.500L = 2M

The equilbrium concentrations are:

[NO] = 2M - 2X

[N₂] = 0.8M +X

[O₂] = X

Replacing:

5.93 = [2 - 2X]² / [0.8+X] [X]

5.93 = 4 - 8X + 4X² / 0.8X + X²

4.744X + 5.93X² = 4 - 8X + 4X²

1.93X² + 12.744X - 4 = 0

Solving for X:

X = -6.9M → False solution. There are no negative concentrations

X = 0.3M. Real solution.

[N₂] in equilibrium is:

[N₂] = 0.8M +0.3M

[N₂] = 1.1M

5 0
3 years ago
134. How many of the following statements are true concerning chemical equations? I. Coefficients can be fractions. II. Subscrip
Katen [24]

Answer: Only I and V are true statements concerning chemical equations

Explanation: II subscripts cannot be fractions and coefficients represent the number of moles of reactants and/or products. Subscripts cannot be changed but coefficients can.

4 0
3 years ago
Read 2 more answers
Equal molar quantities of Ca2 and EDTA (H4Y) are added to make a 0.010 M solution of CaY2- at pH 10. The formation constant for
abruzzese [7]

Answer:

the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

Explanation:

Given the data in the question;

Ca^{2+ + y^{4- ⇄  CaY^{2-

Formation constant Kf

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) = 5.0 × 10¹⁰

Now,

[y^{4-] = \alpha _4CH_4Y; ∝₄ = 0.35

so the equilibrium is;

Ca^{2+ + H_4Y ⇄  CaY^{2- + 4H⁺

Given that; CH_4Y = Ca^{2+     { 1 mol Ca^{2+  reacts with 1 mol H_4Y  }

so at equilibrium, CH_4Y = Ca^{2+ = x

∴

Ca^{2+ + y^{4- ⇄  CaY^{2-

x        + x         0.010-x

since Kf is high, them x will be small so, 0.010-x is approximately 0.010

so;

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) =  CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ 0.010 / ( [x][ 0.35 × x] )  = 5.0 × 10¹⁰

⇒ 0.010 / 0.35x²  = 5.0 × 10¹⁰

⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 5.7142857 × 10⁻¹³

⇒ x = √(5.7142857 × 10⁻¹³)

⇒ x = 7.559 × 10⁻⁷

Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

8 0
3 years ago
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