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3 years ago

Ok here is the next one I’m counting on you Thanks.

Chemistry

1 answer:

kodGreya [7K]3 years ago

7 0

food moves through the mouth by muscle movement

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Whats the correct order of these?

Igoryamba

1109, 5.9 * 10 squared, -0.041, -4.2 times 10 to the -3, and -7.6 times 10 to the -5
Hope this helps

3 0

3 years ago

What was technicalagy like in the past

V125BC [204]

Most of it was analog

8 0

2 years ago

Read 2 more answers

3. Students measured the mass of the reactants and products for a combustion reaction they observed.

coldgirl [10]

Answer:

A. Students made a measurement error, because ending with more products is impossible.

Explanation:

The law of conversation of matter tells us that in a chemical reaction, matter is never created or destroyed, it's simply converted from one form to another. So the mass of reactants should always equal the mass of the products in a chemical reaction. If there is excess mass in the product, the students have made an error of some kind.

4 0

3 years ago

Finding energy and wavelength?

deff fn [24]

The equation is
W = C/F
W= 3.00 x 10^8 m/sec
——————————
6.165 x 10^14 Hz
W= 4.87 x 10^-7 m

Energy is
E=hF
E= (6.626 x 10^-34 Jxsec )(6.165 x 10^14 Hz)
E= 4.085 x 10^-19 J

3 0

3 years ago

Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simulta

scoray [572]

Answer:

70.77 g/mol is the molar mass of the unknown gas.

Explanation:

Effusion is defined as rate of change of volume with respect to time.

Rate of Effusion= $\frac{Volume}{Time}$

Effusion rate of oxygen gas after time t = $E=\frac{4.64 mL}{t}$

Molar mass of oxygen gas = M = 32 g/mol

Effusion rate of unknown gas after time t = $E'=\frac{3.12 mL}{t}$

Molar mass of unknown gas = M'

The rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

$\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

$\frac{E}{E'}=\sqrt{\frac{M'}{M}}$

$\frac{\frac{4.64 mL}{t}}{\frac{3.12 mL}{t}}=\sqrt{\frac{M'}{32 g/mol}}$

M' = 70.77 g/mol

70.77 g/mol is the molar mass of the unknown gas.

8 0

3 years ago