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3 years ago

Neutrons add only ____ to an atom.

2 answers:

8 0

Neutrons are neutral particles, so they don't change the charge of an atom. They do however, change the atomic mass, so the answer would be a. mass.

blagie [28]3 years ago

3 0

the answer for the will be A.mass

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When a cold pack is used, ammonium chloride mixes with water and energy is absorbed from the water to reduce the temperature of

Leto [7]

I think the answer is Endothermic. Since the question says "absorbs," endo means to "go in" sort of and exo means to exit.

exo = exit
endo = enter
(A trick my teacher taught us)

Sorry if I'm wrong, hope I helped. :)

7 0

3 years ago

Read 2 more answers

1. Write a balanced chemical equation for each of the following reactions. Remember that gases of elements such as oxygen are di

Anon25 [30]

Answer:

$CH_4_{(g)}+2O_2_{(g)}\rightarrow CO_2_{(g)}+2H_2O_{(g)}$

$2C_4H_{10}_{(g)}+9O_2_{(g)}\rightarrow 8CO_2_{(g)}+10H_2O_{(g)}$

$H_2SO_4_{(aq)}+2KOH_{(aq)}\rightarrow K_2SO_4_{(aq)}+2H_2O_{(g)}$

Explanation:

(a)

The balanced chemical equation for the reaction of methane with oxygen gas to form carbon dioxide and water is shown below as;

$CH_4_{(g)}+2O_2_{(g)}\rightarrow CO_2_{(g)}+2H_2O_{(g)}$

(b)

The balanced chemical equation for the reaction of butane with oxygen gas to form carbon dioxide and water is shown below as;

$2C_4H_{10}_{(g)}+9O_2_{(g)}\rightarrow 8CO_2_{(g)}+10H_2O_{(g)}$

(c)

The balanced chemical equation for the reaction of solution of sulfuric acid reacts with aqueous potassium hydroxide to produce potassium sulfate and water is shown below as;

$H_2SO_4_{(aq)}+2KOH_{(aq)}\rightarrow K_2SO_4_{(aq)}+2H_2O_{(g)}$

5 0

4 years ago

A chemical reaction that is expected to form 325.0 gof product only forms 123.8 g of product. What is the percent yield of this

Tasya [4]

Answer:

$\boxed {\boxed {\sf A. \ 38.1 \%}}$

Explanation:

Percent yield is the ratio of the amount actually produced to how much could theoretically be produced. It is found using this formula:

$\% \ yield = \frac{actual \ yield}{theoretical \ yield} *100$

For this reaction, the theoretical or expected yield is 325.0 grams. The actual yield is 123.8 grams.

$\% \ yield = \frac{ 123.8 \ g }{325.0 \ g }*100$

Divide.

$\% \ yield = 0.380923077 *100$

$\% \ yield = 38.0923077$

Round to the nearest hundredth. The 9 in the hundredth place tells us to round the 0 to a 1 .

$\% \ yield \approx 38.1$

The percent yield is about <u>38.1%</u>

5 0

3 years ago

Read 2 more answers

Which of these substances are most likely crystalline solids? Select all that apply.

Lostsunrise [7]

Answer:

you did not show the options

Explanation:

8 0

3 years ago

Read 2 more answers

Find the mass of 3.27 x 10^23 molecules of H2SO4. Use 3 significant digits<br> and put the units.

marta [7]

Answer:

Approximately $53.3\; \rm g$ .

Explanation:

Lookup Avogadro's Number: $N_{\rm A} = 6.02\times 10^{23}\; \rm mol^{-1}$ (three significant figures.)

Lookup the relative atomic mass of $\rm H$ , $\rm S$ , and $\rm O$ on a modern periodic table:

$\rm H$ : $1.008$ .
$\rm S$ : $32.06$ .
$\rm O$ : $15.999$ .

(For example, the relative atomic mass of $\rm H$ is $1.008$ means that the mass of one mole of $\rm H\!$ atoms would be approximately $1.008\!$ grams on average.)

The question counted the number of $\rm H_2SO_4$ molecules without using any unit. Avogadro's Number $N_{\rm A}$ helps convert the unit of that count to moles.

Each mole of $\rm H_2SO_4$ molecules includes exactly $(1\; {\rm mol} \times N_\text{A}) \approx 6.02\times 10^{23}$ of these $\rm H_2SO_4 \!$ molecules.

$3.27 \times 10^{23}$ $\rm H_2SO_4$ molecules would correspond to $\displaystyle n = \frac{N}{N_{\rm A}} \approx \frac{3.27 \times 10^{23}}{6.02 \times 10^{23}\; \rm mol^{-1}} \approx 0.541389\; \rm mol$ of such molecules.

(Keep more significant figures than required during intermediary steps.)

The formula mass of $\rm H_2SO_4$ gives the mass of each mole of $\rm H_2SO_4\!$ molecules. The value of the formula mass could be calculated using the relative atomic mass of each element:

$\begin{aligned}& M({\rm H_2SO_4}) \\ &= (2 \times 1.008 + 32.06 + 4 \times 15.999)\; \rm g \cdot mol^{-1} \\ &= 98.702\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the mass of approximately $0.541389\; \rm mol$ of $\rm H_2SO_4$ :

$\begin{aligned}m &= n \cdot M \\ &\approx 0.541389\; \rm mol \times 98.702\; \rm g \cdot mol^{-1}\\ &\approx 53.3\; \rm g\end{aligned}$ .

(Rounded to three significant figures.)

6 0

3 years ago