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4 years ago

HOW DO I FIND ACCELERATION

Chemistry

2 answers:

Stells [14]4 years ago

6 0

Explanation:

Acceleration is given by the formula :

a = (v - u)/t

a is the acceleration , u is the initial velocity , v is the final velocity , t is the time in seconds

agasfer [191]4 years ago

4 0

Answer: acceleration= Change in velocity/ change in time

Explanation:

A= Delta v/ delta t

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A piece of unknown metal weighs 348g. When the metal piece absorbs 6.64kj of heat , its temperature increases from 24.4C to 43.6

Morgarella [4.7K]

Answer:

This metal could be the aluminium with a specific heat of $H = 993 \frac{J}{kgC}$

Explanation:

A pie of unknown metal presents a mass (M) of 348 g. This metal is heated using energy (E) of 6.64 kJ and the temperature increases from T1 =24.4 to T2 =43.6°C. We can calculate the specific heat (H) of this metal as follows

$H = \frac{E}{M*(T2-T1)}$

We can replace previously presented data in this equation. After simplifying and converting to adequated units, we found that

$H = \frac{6640 J}{0.348 Kg*(43.6-24.4) C} =\frac{6640 J}{6.686 KgC}$

Finally, the specific heat of this metal is

$H = 993 \frac{J}{kgC}$

The aluminium could be the metal, its specific heat is similar to that found in this problem.

Finally, we can conclude that this metal could be the aluminium with a specific heat of $H = 993 \frac{J}{kgC}$

7 0

3 years ago

A 25.0-mL sample of 0.150 M hydrocyanic acid is titrated with a 0.150 M NaOH solution. The Ka of hydrocyanic acid is 4.9 × 10-10

lara [203]

Answer:

The pOH = 1.83

Explanation:

Step 1: Data given

volume of the sample = 25.0 mL

Molarity of hydrocyanic acid = 0.150 M

Molarity of NaOH = 0.150 M

Ka of hydrocyanic acid = 4.9 * 10^-10

Step 2: The balanced equation

HCN + NaOH → NaCN + H2O

Step 3: Calculate the number of moles hydrocyanic acid (HCN)

Moles HCN = molarity * volume

Moles HCN = 0.150 M * 0.0250 L

Moles HCN = 0.00375 moles

Step 3: Calculate moles NaOH

Moles NaOH = 0.150 M * 0.0305 L

Moles NaOH = 0.004575 moles

Step 4: Calculate the limiting reactant

0.00375 moles HCN will react with 0.004575 moles NaOH

HCN is the limiting reactant. It will completely be reacted. There will react 0.00375 moles NaOH. There will remain 0.004575 - 0.00375 = 0.000825 moles NaOH

Step 5: Calculate molarity of NaOH

Molarity NaOH = moles NaOH / volume

Molarity NaOH = 0.000825 moles / 0.0555 L

Molarity NaOH = 0.0149 M

Step 6: Calculate pOH

pOH = -log [OH-]

pOH = -log (0.0149)

pOH = 1.83

The pOH = 1.83

6 0

3 years ago

What is the volume of 425g of ice if the density is 0.92g/mL

Sati [7]

Density=mass/ volume so you solve for volume and get 461.96 mL

6 0

4 years ago

Consider the formation of [Ni(en)3]2+ from [Ni(H2O)6]2+. The stepwise ΔG∘ values at 298 K are ΔG∘1 for first step=−42.9 kJ⋅mol−1

timurjin [86]

Answer:

kf = 1.16 x 10¹⁸

Explanation:

Step 1: [Ni(H₂O)₆]²⁺ + 1en → [Ni(H₂O)₄(en)]²⁺ ΔG°1 = -42.9 kJmol⁻¹

Step 2: [Ni(H₂O)₄(en)]²⁺ + 1en → [Ni(H₂O)₂(en)₂]²⁺ ΔG°2 = -35.8 kJmol⁻¹

Step 3: [Ni(H₂O)₂(en)₂]²⁺ + 1en → [Ni(en)₃]²⁺ ΔG°3 = -24.3 kJmol⁻¹

________________________________________________________

Overall reaction: [Ni(H₂O)₆]²⁺ + 3en → [Ni(en)₃]²⁺ ΔG°r

ΔG°r = ΔG°1 + ΔG°2 + ΔG°3

ΔG°r = -42.9 - 35.8 - 24.3

ΔG°r = -103.0 kJmol⁻¹

ΔG°r = -RTlnKf

-103,000 Jmol⁻¹ = - 8.31 J.K⁻¹mol⁻¹ x 298 K x lnKf

kf = e ^(-103,000/-8.31x298)

kf = e ^41.59

kf = 1.16 x 10¹⁸

7 0

3 years ago

GT bisects BN at point G.<br><br> If BN = 24 find BG

LenaWriter [7]

Answer:

Explanation:

If GT bisects BN at point G, then point G would be in the middle of BN. So, since Bg and NG have to be equal and add up to 24, then the answer is 12.

4 0

3 years ago