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3 years ago

A satellite image can help scientists locate the area where two plates have diverged by showing the existence of

Chemistry

1 answer:

Alex Ar [27]3 years ago

8 0

Answer:

Explanation:

Because it's d so yeah

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What element provides strength to the exoskeleton of Clams and oysters

Julli [10]

Phosphorus!!!! Hope this helps

8 0

3 years ago

Calculate the moles of calcium oxalate formed from 5 grams of potassium oxalate (molar mass 184.24 g/mol) the mole mole ratio fr

Nataly_w [17]

5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.

Calcium oxalate (CaC₂O₄) is obtained by the reaction of 5 g of potassium oxalate (K₂C₂O₄).

We can calculate the moles of CaC₂O₄ obtained considering the following relationships.

The molar mass of K₂C₂O₄ is 184.24 g/mol.
The mole ratio of K₂C₂O₄ to CaC₂O₄ is 2:1.

$5 g K_2C_2O_4 \times \frac{1molK_2C_2O_4}{184.24gK_2C_2O_4} \times \frac{1molCaC_2O_4}{2molK_2C_2O_4} = 0.03molCaC_2O_4$

5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.

Learn more: brainly.com/question/15288923

7 0

3 years ago

Which of the following is a half-reaction? A. Zn+CuSO4−> B. 2Cl−−>Cl2+2e− C. H2+1/2O2−>H2O D. −>Cu+ZnSO4

malfutka [58]

Answer:

2Cl——>Cl2+2e-

Explanation:

It shows an electron loss or gain

8 0

3 years ago

Which statements are scientific statements

Harrizon [31]

Wym there is no question or statements.

6 0

3 years ago

When 60 mL of 0.22 M NH4Cl is added to 60 mL of 0.22 M NH3, relative to the pH of the 0.10 M NH3 solution the pH of the resultin

densk [106]

Answer:

Will be more acidic

Explanation:

The equilibrium of NH3 in water is:

NH3(aq) + H2O(l) ⇄ NH4⁺(aq) + OH⁻(aq).

Where equilibrium constant, Kb, is:

Kb = 1.85x10⁻⁵ = [NH4⁺] [OH⁻] / [NH3]

From 0.10M NH3, the reaction will produce X of NH4⁺ and X of OH⁻ and Kb will be:

1.85x10⁻⁵ = [X] [X] / [0.10M]

1.8x10⁻⁶ = X²

X = 1.34x10⁻³ = [OH⁻]

As pOH = -log[OH⁻] = 2.87

And as pH = 14 - pOH

pH of the 0.10M NH3 is 11.13

Now, to find the pH of the NH4Cl and NH3 we need to use H-H equation for bases:

pOH = pKb + log [NH4⁺] / [NH3]

<em>Where pKb is -log Kb = 4.74 and [] are moles of both compounds.</em>

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Moles of [NH4⁺] = [NH3] = 60mL, 0.060L*0.22M = 0.0132moles:

pOH = 4.74 + log [0.0132] / [0.0132]

pOH = 4.74

pH = 14 - 4.74 = 9.26

That means the pH of the resulting solution will be more acidic

7 0

3 years ago