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natta225 [31]
3 years ago
14

PLEASE HELP! ITS INTEGRATED SCIENCE

Chemistry
2 answers:
Lady bird [3.3K]3 years ago
7 0

Answer:

-4

16:00

8

it's pretty accurate besides it's way of numbering.

Mrac [35]3 years ago
3 0
Answer :
-4

16:00

8

Accurate and way of numbering
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What will happen if there is an insufficient supply of oxygen during the combustion of a hydrocarbon?
Novay_Z [31]
Then it will blow or make a big eruption then you might die
7 0
4 years ago
A cube has a mass of 42 grams and a volume of 15 cubic centimeters. What is it’s density?
deff fn [24]

Answer:

2.8g/cm³

Explanation:

Given parameters:

Mass of cube = 42g

Volume of cube  = 15cm³

Unknown:

Density of the cube  = ?

Solution:

Density is defined as the mass per unit volume of a substance. It is mathematically expressed as:

 Density  = \frac{mass}{volume}  

So;

 Density  = \frac{42}{15}   = 2.8g/cm³

6 0
3 years ago
What occurs when the nuclei of two very small atoms combine into one large nucleus?
Svetlanka [38]

Answer:

The combination of nuclei to form a bigger and heavier nucleus is known as Fussion. The consequence of fusion is the absorption or release of energy.

7 0
3 years ago
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
Reactions that affect the nucleus of an atom are called ?
densk [106]
I believe the answer is C
4 0
3 years ago
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