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2 years ago

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Martin is conducting an experiment. His first test gives him a yield of 5.2 grams. His second test gives him a yield of 1.3 gram

s. His third test gives him a yield of 8.5 grams. On average, his yield is 5.0 grams, which is close to the known yield of 5.1 grams of substance. Which of the following are true?
A. His results are neither accurate nor precise

B. His results are both accurate and precise

C. His results are accurate but not precise

D. His results are precise but not accurate

1 answer:

Anvisha [2.4K]2 years ago

4 0

I Think The answer is c I hope it helps My friend Message Me if I’m wrong and I’ll change My answer and fix it for you

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If 36.9 mL of B2H6 reacted with excess oxygen gas, determine the actual yield of B2O3 if the percent yield of B2O3 was 75.7%. (T

zhuklara [117]

Answer: The actual yield of $B_2O_3$ is 60.0 g

Explanation:-

The balanced chemical reaction :

$B_2H_6(l)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)$

Mass of $B_2H_6$ = $Density\times Volume=1.131g/ml\times 36.9ml=41.7g$

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} B_2H_6=\frac{41.7g}{27.668g/mol}=1.51moles$

According to stoichiometry:

1 mole of $B_2H_6$ gives = 1 mole of $B_2O_3$

1.51 moles of $B_2H_6$ gives = $\frac{1}{1}\times 1.51=1.51$ moles of $B_2O_3$

Theoretical yield of $B_2O_3=moles\times {Molar mass}}=1.14mol\times 69.62g/mol=79.3g$

Percent yield of $B_2O_3$ = $75.7\%$

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

$75.7\%=\frac{\text{Actual yield}}{79.3}\times 100$

${\text{Actual yield}}=60.0g$

Thus the actual yield of $B_2O_3$ is 60.0 g

7 0

2 years ago

Calculate the mass in grams of 0.800 mole of H2CO3. g

PilotLPTM [1.2K]

MH₂CO₃: (1g×2) + 12g + (16g×3) = 62 g/mol

1 mol --- 62g
0,8 mol -- X
X = 0,8×62
X = 49,6g

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2 years ago

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jin has four colorful glass bottles on his windowsill. one is red, one is violet, one is green, and one is yellow. the light tha

Over [174]

The red bottle would have the lowest frequency because red light has the longest wavelengths. The light passing through the violet would have the highest frequency because its wavelengths are the shortest.

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2 years ago

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. Citric acid, which can be obtained from lemon juice, has the molecular formula C6H8O7. A 0.250-g sample of citric acid dissolv

marysya [2.9K]

Answer:

3 acidic hydrogens per molecule of citric acid

Explanation:

In a sample of 37.2 mL(0.0372 L) of 0.105 mol/L of NaOH, will have:

n = 0.105x0.0372 = 0.0039 mol of NaOH

The dissociation of NaOH will give the same number of moles of Na⁺ and OH⁻.

The molar mass of citric acid is:

C: 12g/mol x 6 = 72 g/mol

H: 1g/mol x 8 = 8g/mol

O: 16 g/mol x 7 = 112 g/mol

192 g/mol

So, 0.250g of the acid has

n = mass/molar mass

n = 0.250/192

n = 0.0013 mol.

To be neutralized, it will be necessary 0.0039 mol of acidic hydrogens to react with the 0.0039 mol of OH⁻.

The dissociation reaction of one molecule of the acid will give the stoichiometry:

1 mol of acid ----------------------- x mol of acidic hydrogens

0.0013 mol --------------------------- 0.0039

For a simple direct three rule:

0.0013x = 0.0039

x = 3 acidic hydrogens per molecule of citric acid.

3 0

2 years ago

Coal gasification is a multistep process to convert coal into cleaner-burning fuels. In one step, a coal sample reacts with supe

ddd [48]

Answer :

The enthalpy of reaction is, -187.6 kJ/mol

The total heat will be, -2251 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

(a) The formation of $CH_4$ will be,

$2C(coal)+2H_2O(g)\rightarrow CH_4(g)+CO_2(g)$ $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $C(coal)+H_2O(g)\rightarrow CO(g)+H_2(g)$ $\Delta H_1=29.7kJ$

(2) $CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)$ $\Delta H_2=-41kJ$

(3) $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$ $\Delta H_3=-206kJ$

We are multiplying equation 1 by 2 and then adding all the equations, we get :

(b) The expression for enthalpy of reaction will be,

$\Delta H_{rxn}=2\times \Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H_{rxn}=(2\times 29.7)+(-41)+(-206)$

$\Delta H_{rxn}=-187.6kJ/mol$

Therefore, the enthalpy of reaction is, -187.6 kJ/mol

(c) Now we have to calculate the total heat.

$\Delta H=\frac{q}{n}$

or,

$q=\Delta H\times n$

where,

$\Delta H$ = enthalpy change = -187.6 kJ/mol

q = heat = ?

n = number of moles of coal = $\frac{1.00\times 1000g}{12.00g/mol}=83.33mol$

Now put all the given values in the above formula, we get:

$q=(-187.6kJ/mol)\times (83.33mol)=-2.251kJ$

Thus, the total heat will be, -2251 kJ

4 0

2 years ago