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Vesna [10]
3 years ago
8

Help please i am very dumb

Chemistry
1 answer:
UkoKoshka [18]3 years ago
7 0

Answer:

1. 25%

2. 1.25

3. 1

Explanation:

Be sure to look at the x and y axis to answer these questions. All you need to do is look at the graph.

- Hope that helps! Please let me know if you need further explanation.

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A 52.0 g of Copper (specific heat=0.0923cal/gC) at 25.0C is warmed by the addition of 299 calories of energy. find the final tem
Leto [7]

Answer : The final temperature of the copper is, 87.29^oC

Solution :

Formula used :

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

where,

Q = heat gained  = 299 cal

m = mass of copper = 52 g

c = specific heat of copper = 0.0923cal/g^oC      

\Delta T=\text{Change in temperature} 

T_{final} = final temperature = ?

T_{initial} = initial temperature = 25^oC

Now put all the given values in the above formula, we get the final temperature of copper.

299cal=52g\times 0.0923cal/g^oC\times (T_{final}-25^oC)

T_{final}=87.29^oC

Therefore, the final temperature of the copper is, 87.29^oC

4 0
4 years ago
What are the products of photosynthesis
spin [16.1K]
Oxygen (6O2) and Glucose (C6H12O6)

<span>Reference: 6CO2 + 6H2O + light energy = C6H12O6 + 6O2.</span>
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Which metals or non-metals are liquid at a room temperature of 25°C?​
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Answer:

gallium

Explanation:

it is the old melting in a hot cup of coffee spoon

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5. for the following measurements is expressed to three significant figures? a) 0.007 m b) 7077 mg c) 7.30 x 10^-7 km d) 0.070
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The answer is c
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3 years ago
Find the initial concentration of the weak acid or base in each of the following aqueous solutions: (a) a solution of HClO with
Luda [366]

Answer:

a) 0.021 M

b) 0.019 M

Explanation:

To do this, you need to calculate the concentration of ions in solution with the given value of pH for each solution, then, write the chemical equation for both solutions, Set an ICE chart, use the value of Ka and Kb reported for both solutions, and solve for the initial concentration.

This is the general procedure to do it, now let's do it by parts.

<em><u>a) Concentration of HClO pH = 4.6</u></em>

With the given pH, we use the following expression:

pH = -log[H₃O⁺]      From here, we solve for [H₃O⁺]

[H₃O⁺] = 10^(-pH)   (1)

Let's calculate first the hydronium concentration:

[H₃O⁺] = 10^(-4.6) = 2.51x10⁻⁵ M

This value indicates the equilibrium concentration of this ion in solution. Now, to know the initial concentration of the acid, we need to do an ICE chart and write the chemical equation. This is an acid - base reaction, so we need the value of Ka of the acid.

         HClO + H₂O <---------> H₃O⁺ + ClO⁻       Ka = 3x10⁻⁸

I:            Y                                 0          0

C:          -x                                +x         +x

E:           Y - x                            x          x

With this chart, we need to write the expression for Ka which is:

Ka = [H₃O⁺] * [ClO⁻] / [HClO] = x² / Y-x

But we already know the concentration of [H₃O⁺], which is the same for [ClO⁻], and the value of Ka, so all we have to do is replace the values in the above expression and solve for Y:

3x10⁻⁸ = (2.51x10⁻⁵)² / Y - 2.51x10⁻⁵

We can round to Y because "x" is a very small value as it's value of Ka so:

3x10⁻⁸ = (2.51x10⁻⁵)²/Y

Y = (2.51x10⁻⁵)²/3x10⁻⁸

<h2><em>Y = [HClO] = 0.021 M</em></h2>

<em>And this is the initial concentration of the acid.</em>

<u><em>b) Solution of hidrazine pH = 10.2</em></u>

We do the same procedure as part a) with the difference that instead of using Ka , we use Kb and concentration of [OH⁻]. The Kb for hydrazine is 1.3x10⁻⁶

Let's calculate the [OH⁻]:

pOH = 14 - pH

pOH = 14 - 10.2 = 3.8

[OH⁻] = 10^(-3.8) = 1.58x10⁻⁴ M

The chemical equation:

          N₂H₄ + H₂O <---------> N₂H₅⁺ + OH⁻    Kb = 1.3x10⁻⁶

I:            Y                                  0           0

C:          -x                                +x           +x

E:         Y-x                                 x           x

Kb = x²/(Y-x)

1.3x10⁻⁶ = (1.58x10⁻⁴)²/Y

Y = (1.58x10⁻⁴)²/1.3x10⁻⁶

<h2><em><u>Y = [OH⁻] = 0.019 M</u></em></h2>

And this is the initial concentration of hydrazine

4 0
3 years ago
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