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2 years ago

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Which of the following statements is true? A. Isotopes of the same element have the same number of protons and neutrons. B. Isot

opes have the same physical properties as the normal atom but different chemical properties. C. Isotopes have the same chemical properties as the normal atom but different physical properties. D. If an isotope of one element has the same atomic mass as another element, they will have the same properties.

1 answer:

k0ka [10]2 years ago

6 0

Answer:

B

Explanation:

Isotopes have the same number of protons but different number of neutrons.

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What is thr valency of scadium

Hoochie [10]

scandium is a transition metal and shows a valency of 3.

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3 years ago

What is the total pressure of a container in kpa if the partial pressures are 1.25 atm and 66.7 psi?

STatiana [176]

Answer:

586 kpa(kilopascal/1000 pascals)

Explanation:

given 1.24 atm(standard atmosphere), and 66.7 psi(pound force per square inch).

To find the total pressure we should use dalton's law of partial pressures which is the sum of the pressures of each individual gas.

then we convert them to pascals and divide by 1000 to get the measurement in kilopascal.

knowing that 1 atmosphere is proportional to around 14.696 psi. We can multiply our given measure of atm by that and sum it by psi like so. 1.24×14.6959 = 18.22298.

Then,

18.22298+ 66.7 = 84.92298

psi.

Since 1 psi is proportional to around 6894.76 pascals. 1 psi will be 68.9476 kilopascal. 84.92298 * 6.89476 = 585.523336 ≈ 586

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2 years ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

2 years ago

What is the volume of 0.120 g of C2H2F4<br> vapor at 0.970 atm and 22.4°C?<br> Answer in units of L.

ludmilkaskok [199]

Answer:

V= 0.031L

Explanation:

P= 0.97atm, V= ?, n= 0.12/98 =0.00122mol, R= 0.082, T= 22.4+273= 295.4

Applying

PV=nRT

0.970×V = 0.00122×0.082×295.4

Simplify the above equation

V= 0.031L

6 0

2 years ago

Values for the molar mass of nitrogen, oxygen, and nitrogen dioxide

CaHeK987 [17]

Answer:

n=N/Na

n = \frac{8.23 \times {10}^{22} }{6.02 \times {10}^{23} } = 0.1367 \: mol

answer: 0.14 mol

Explanation:

5 0

2 years ago