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3 years ago

if the density of a bar of gold is 19.3g/cm^3and you cut it into four pieces , what is its density of each piece of gold

1 answer:

4 0

Density is an intrinsic property, so it is independent of the amount of substance present: one gold coin would have the same density as a solid gold boulder.

So if the density of gold is 19.3 g/cm³, the density of a bar of gold and the pieces into which the bar is cut would all be 19.3 g/cm³.

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(5.625 + 8.15) x 2.34 + 3.2

dalvyx [7]

The answer is 35.4335

Hope this helped! (Plz mark me brainliest!)

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3 years ago

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If a reaction has an equilibrium constant slightly greater than 1, what type of reaction is it?

Paha777 [63]

Reversible, flavoring products

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3 years ago

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Write the full ionic equation and net ionic equation for sodium dihydrogen phosphate + calcium carbonate, sodium oxilate + calcl

My name is Ann [436]

Answer:

Sodium dihydrogen phosphate + calcium carbonate

Full ionic equation

2 Na⁺(aq) + 2 H₂PO₄⁻(aq) + CaCO₃(s) ⇄ 2 Na⁺(aq) + CO₃²⁻(aq) + Ca(H₂PO₄)₂(s)

Net ionic equation

2 H₂PO₄⁻(aq) + CaCO₃(s) ⇄ CO₃²⁻(aq) + Ca(H₂PO₄)₂(s)

Sodium oxalate + calcium carbonate

Full ionic equation

2 Na⁺(aq) + C₂O₄²⁻(aq) + CaCO₃(s) ⇄ 2 Na⁺(aq) + CO₃²⁻(aq) + CaC₂O₄(s)

Net ionic equation

C₂O₄²⁻(aq) + CaCO₃(s) ⇄ CO₃²⁻(aq) + CaC₂O₄(s)

Sodium hydrogen phosphate + calcium carbonate

Full ionic equation

2 Na⁺(aq) + HPO₄²⁻(aq) + CaCO₃(s) ⇄ CaHPO₄(s) + 2 Na⁺(aq) + CO₃²⁻(aq)

Net ionic equation

HPO₄²⁻(aq) + CaCO₃(s) ⇄ CaHPO₄(s) + CO₃²⁻(aq)

Explanation:

Let's consider two kind of equations:

Full ionic equation: includes all ions and species that do not dissociate in water.
Net ionic equation: includes only ions that participate in the reaction (not spectator ions) and species that do not dissociate in water.

4 0

3 years ago

Which of these formulas is NOT consistent with the guidelines for creating ionic compounds? Mark all that apply.

iVinArrow [24]

Answer:

Explanation:

because boron and fluorine are both nonmetals and don't fit the guidlelines for creating ionic compounds

8 0

3 years ago

The rate of disappearance of HBr in the gas phase reaction2HBr(g) → H2(g) + Br2(g)is 0.301 M s-1 at 150°C. The rate of appearanc

jok3333 [9.3K]

Answer: 0.151

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$2HBr(g)\rightarrow H_2(g)+Br_2(g$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

$Rate=-\frac{1d[HBr]}{2dt}=+\frac{1d[H_2]}{2dt}=+\frac{1d[Br_2]}{dt}$

Given: $-\frac{d[HBr]}{dt}]=0.301$

Putting in the values we get:

$\frac{0.301}{2}=+\frac{1d[Br_2]}{dt}$

$+\frac{1d[Br_2]}{dt}=0.151$

Thus the rate of appearance of $Br_2$ is 0.151

8 0

3 years ago