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2 years ago

Who can teach me chemical counting? please...

Chemistry

1 answer:

kupik [55]2 years ago

4 0

Explanation:

sorry hindi ko alam eii!!!!

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LOOK AT THE IMAGE ABOVE PLEASE SOMEONE HELP OMG PLEASE IM LITERALLY STRESSED PLEASE HELP. I got the answer wrong. Please help an

Anna35 [415]

D, the more liquid there is, the less the temperature will be affected

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3 years ago

How many micrograms (ug) are in 3.4 x 10^-5 ounces (oz)

marysya [2.9K]

Answer:

964ug

Explanation:

The problem here involves converting from one unit to another.

We are to convert from ounces to micrograms.

1ug = 1 x 10⁻⁶g

1oz = 28.35g

So we first convert to grams from oz then take to ug:

Solving:

1oz = 28.35g

3.4 x 10⁻⁵oz will then give 3.4 x 10⁻⁵ x 28.35 = 9.64 x 10⁻⁴g

So;

1 x 10⁻⁶g = 1ug

9.64 x 10⁻⁴g will give $\frac{9.64 x 10^{-4} }{1 x 10^{-6} }$ = 9.64 x 10²ug or 964ug

8 0

2 years ago

What do the colors produced by fireworks indicate?

ahrayia [7]

<span>Metal compounds are used since metals have characteristic flame colors </span>

5 0

3 years ago

Many portable gas heaters and grills use propane, C3H8(g). Using enthalpies of formation, calculate the quantity of heat produce

Elodia [21]

Answer: 600 kJ

Explanation:

C₃H₈ (g) + 5 O₂ (g) =============== 3 CO₂ (g) + 4 H₂O (l)

Δ⁰Hf kJ/mol -104 0 -393.5 -285.8

Δ⁰Hcomb C₃H₈ = 3(-393.5) + 4 (-285.80) - (-104) kJ/mol

Δ⁰Hcomb = 2219.70 kJ/mol

n= m /MW MW c₃H₈ = 44.1 g/mol

n= 12 g/44.1 g/mol = 0.27 mol

then for 12 g the heat released will be

0.27 mol x 2219.70 kJ/mol = 600 KJ

4 0

2 years ago

Read 2 more answers

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

Who can teach me chemical counting? please...​

Who can teach me chemical counting? please...