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3 years ago

trial 2, 2.68 g/cm/3 trial 3, 2.84g/cm/3. aluminum has a density of 2.70g/cm/3 calcute the percent error for each trial.

Chemistry

1 answer:

serg [7]3 years ago

6 0

Answer:

trial 2: 0.74

trial 3: 5.19

I think but the equation for sloving percent error is:

(true value - determined value)/true value * 100

Explanation:

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A sample of ammonia ^NH3h gas is completely decomposed to nitrogen and hydrogen gases over heated iron wool. If the total pressu

icang [17]

Answer : The partial pressure of $N_2$ and $H_2$ is, 216.5 mmHg and 649.5 mmHg

Explanation :

According to the Dalton's Law, the partial pressure exerted by component 'i' in a gas mixture is equal to the product of the mole fraction of the component and the total pressure.

Formula used :

$p_i=X_i\times p_T$

$X_i=\frac{n_i}{n_T}$

So,

$p_i=\frac{n_i}{n_T}\times p_T$

where,

$p_i$ = partial pressure of gas

$X_i$ = mole fraction of gas

$p_T$ = total pressure of gas

$n_i$ = moles of gas

$n_T$ = total moles of gas

The balanced decomposition of ammonia reaction will be:

$2NH_3\rightarrow N_2+3H_2$

Now we have to determine the partial pressure of $N_2$ and $H_2$

$p_{N_2}=\frac{n_{N_2}}{n_T}\times p_T$

Given:

$n_{N_2}=1\\\\n_{H_2}=3\\\\n_{T}=4\\\\p_T=866mmHg$

$p_{N_2}=\frac{1}{4}\times (866mmHg)=216.5mmHg$

and,

$p_{H_2}=\frac{n_{H_2}}{n_T}\times p_T$

Given:

$n_{H_2}=1\\\\n_{H_2}=3\\\\n_{T}=4\\\\p_T=866mmHg$

$p_{H_2}=\frac{3}{4}\times (866mmHg)=649.5mmHg$

Thus, the partial pressure of $N_2$ and $H_2$ is, 216.5 mmHg and 649.5 mmHg

5 0

3 years ago

For the wild type (unmutated) enzyme, you measure a rate of p-nitrophenol release by the change in absorbance at 405 nm (for the

Aleks04 [339]

Answer:

1.2x10⁻⁵M = Concentration of the product released

Explanation:

Lambert-Beer's law states the absorbance of a solution is directly proportional to its concentration. The equation is:

A = E*b*C

Where A is the absotbance of the solution: 0.216

E is the extinction coefficient = 18000M⁻¹cm⁻¹

b is patelength = 1cm

C is concentration of the solution

Replacing:

0.216 = 18000M⁻¹cm⁻¹*1cm*C

<h3>1.2x10⁻⁵M = Concentration of the product released</h3>

4 0

3 years ago

When barium hydroxide and ammonium chloride react,the temperature of the mixture decreases.What kind of reaction is this?

GalinKa [24]

Answer:

Endothermic

Explanation:

7 0

3 years ago

A buffer solution is composed of 4.00 4.00 mol of acid and 3.25 3.25 mol of the conjugate base. If the p K a pKa of the acid is

Reika [66]

Answer: The pH of the buffer is 4.61

Explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{conjuagate base}]}{[\text{acid}]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of weak acid = 4.70

$[\text{conjuagate base}]}$ = moles of conjugate base = 3.25 moles

$[\text{acid}]$ = Moles of acid = 4.00 moles

pH = ?

Putting values in above equation, we get:

$pH=4.70+\log(\frac{3.25}{4.00})\\\\pH=4.61$

Hence, the pH of the buffer is 4.61

8 0

3 years ago

HELP PLEASE! Picture for question provided

BartSMP [9]

Answer:

Explanation:

The first one is CrO. The Chromium has the same charge as the oxygen so mol numbers are dropped.

The Second one is CrO2 The two oxygens have a charge of 2(-2) = -4. To balance this, the Chromium must have a charge of +4 Cr(Iv)O2

The third one is can be set up like this

Cr + 3(-2) = 0

Cr - 6 = 0

Cr = 6

Therefore the formula is Cr(vi)O3

The last one is a bit tricky. Follow this carefully. There are 2 Crs and 3Os.

The formula looks like this

2Cr + 3(-2) = 0

2Cr - 6 = 0

2Cr = 6

Cr = 3

The formula is Cr(iii)2 O3

5 0

2 years ago